{"id":1118,"date":"2016-02-16T18:34:13","date_gmt":"2016-02-16T18:34:13","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&#038;p=1118"},"modified":"2018-01-04T00:01:47","modified_gmt":"2018-01-04T00:01:47","slug":"7-3-1-solving-radical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/chapter\/7-3-1-solving-radical-equations\/","title":{"raw":"Solving Radical Equations","rendered":"Solving Radical Equations"},"content":{"raw":"&nbsp;\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve Radical Equations\r\n<ul>\r\n \t<li>Isolate square roots in equations and solve for a variable<\/li>\r\n \t<li>Identify extraneous solutions to radical equations<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Square Roots and Completing the Square for Solving Radical Equations\r\n<ul>\r\n \t<li>Use square roots to solve quadratic\u00a0equations<\/li>\r\n \t<li>Complete the square to solve a quadratic equation<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Using the Quadratic Formula to Solve Quadratic Equations\r\n<ul>\r\n \t<li>Write a quadratic equation in standard form and identify the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> in a\u00a0standard form quadratic equation.<\/li>\r\n \t<li>Use the Quadratic Formula to find all real solutions.<\/li>\r\n \t<li>Solve application problems requiring the use of the Quadratic Formula.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nA basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations\u2014rearrange the expression to isolate the variable you want to know, and then solve the resulting equation.\r\n<div class=\"textbox shaded\">\r\n<h4>Solutions to Radical Equations<\/h4>\r\nThe solutions of [latex]x^2=a[\/latex] are called the square roots of a.\r\n<ul>\r\n \t<li>When a is positive, a &gt; 0, [latex]x^2=a[\/latex] has two solutions, [latex]+\\sqrt{a},-\\sqrt{a}[\/latex]. [latex]+\\sqrt{a}[\/latex] is the nonnegative square root of a, and [latex]-\\sqrt{a}[\/latex] is the negative square root of a.<\/li>\r\n \t<li>When a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions.<\/li>\r\n \t<li>When a is zero, a = 0, [latex]x^2=a[\/latex] has one solution: a = 0<\/li>\r\n<\/ul>\r\n<\/div>\r\nJust to drive home the importance of the concept that when a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions, we will restate it in words. \u00a0If you have a negative number under a square root sign as in this example,\r\n\r\n[latex]\\sqrt{-3}[\/latex]\r\n\r\nThere will be no real number solutions.\r\n\r\nThere are two key ideas that you will be using to solve radical equations. The first is that if [latex] a=b[\/latex], then [latex] {{a}^{2}}={{b}^{2}}[\/latex]. (This property allows you to square both sides of an equation and remain certain that the two sides are still equal.) The second is that if the square root of any nonnegative number <i>x<\/i> is squared, then you get <i>x<\/i>: [latex] {{\\left( \\sqrt{x} \\right)}^{2}}=x[\/latex]. (This property allows you to \u201cremove\u201d the radicals from your equations.)\r\n\r\nLet\u2019s start with a radical equation that you can solve in a few steps:[latex] \\sqrt{x}-3=5[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] \\sqrt{x}-3=5[\/latex]\r\n\r\n[reveal-answer q=\"946356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"946356\"]Add 3 to both sides to isolate the variable term on the left side of the equation.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\r\nCollect like terms.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{x}=8[\/latex]<\/p>\r\nSquare both sides to remove the radical, since [latex] {{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the 8 also! Then simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=64[\/latex] is the solution to [latex] \\sqrt{x}-3=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo check your solution, you can substitute 64 in for <i>x<\/i> in the original equation. Does [latex] \\sqrt{64}-3=5[\/latex]? Yes\u2014the square root of 64 is 8, and [latex]8\u22123=5[\/latex].\r\n\r\nNotice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.\r\n\r\nIn the example above, only the variable <i>x<\/i> was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>. Watch how the next two problems are solved.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] \\sqrt{x+8}=3[\/latex]\r\n[reveal-answer q=\"673245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673245\"]Notice how the radical contains a binomial: [latex]x+8[\/latex]. Square both sides to remove the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{x+8} \\right)}^{2}}={{\\left( 3 \\right)}^{2}}[\/latex]<\/p>\r\n[latex] {{\\left( \\sqrt{x+8} \\right)}^{2}}=x+8[\/latex]. Now simplify the equation and solve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}x+8=9\\\\x=1\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting 1 for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{1+8}=3\\\\\\sqrt{9}=3\\\\3=3\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=1[\/latex] is the solution to [latex] \\sqrt{x+8}=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show how to solve simple radical equations.\r\nhttps:\/\/youtu.be\/tT0Zwsto6AQ\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] 1+\\sqrt{2x+3}=6[\/latex]\r\n\r\n[reveal-answer q=\"479262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479262\"]Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSimplify the equation and solve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting 11 for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=11[\/latex] is the solution for [latex] 1+\\sqrt{2x+3}=6[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Solving Radical Equations<\/h3>\r\nFollow the following four steps to solve radical equations.\r\n<ol>\r\n \t<li>Isolate the radical expression.<\/li>\r\n \t<li>Square both sides of the equation: If [latex]x=y[\/latex] then [latex]x^{2}=y^{2}[\/latex].<\/li>\r\n \t<li>Once the radical is removed, solve for the unknown.<\/li>\r\n \t<li>Check all answers.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2>Identify Extraneous\u00a0Solutions<\/h2>\r\nFollowing rules is important, but so is paying attention to the math in front of you\u2014especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] \\sqrt{a-5}=-2[\/latex]\r\n\r\n[reveal-answer q=\"798652\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798652\"]Square both sides to remove the term [latex]a\u20135[\/latex] from the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{a-5} \\right)}^{2}}={{(-2)}^{2}}[\/latex]<\/p>\r\nWrite the simplified equation, and solve for <i>a<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}a-5=4\\\\a=9\\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nNow check the solution by substituting [latex]a=9[\/latex] into the original equation.\r\n\r\nIt does not check!\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{9-5}=-2\\\\\\sqrt{4}=-2\\\\2\\ne -2\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nNo solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLook at that\u2014the answer [latex]a=9[\/latex] does not produce a true statement when substituted back into the original equation. What happened?\r\n\r\nCheck the original problem: [latex]\\sqrt{a-5}=-2[\/latex]. Notice that the radical is set equal to [latex]\u22122[\/latex], and recall that the principal square root of a number can only be <i>positive<\/i>. This means that no value for <i>a<\/i> will result in a radical expression whose positive square root is [latex]\u22122[\/latex]! You might have noticed that right away and concluded that there were no solutions for <i>a<\/i>.\r\n\r\nIncorrect values of the variable, such as those that are introduced as a result of the squaring process are called <strong>extraneous solutions<\/strong>. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important\u2014if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.\r\nIn the next video example, we solve more radical equations that may have extraneous solutions.\r\nhttps:\/\/youtu.be\/qkZHKK77grM\r\n\r\nHave a look at the following problem. Notice how the original problem is [latex] x+4=\\sqrt{x+10}[\/latex], but after both sides are squared, it becomes [latex] {{x}^{2}}+8x+16=x+10[\/latex]. Squaring both sides may have introduced an extraneous solution.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] x+4=\\sqrt{x+10}[\/latex]\r\n\r\n[reveal-answer q=\"705028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705028\"]Square both sides to remove the term [latex]x+10[\/latex]\u00a0from the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( x+4 \\right)}^{2}}={{\\left( \\sqrt{x+10} \\right)}^{2}}[\/latex]<\/p>\r\nNow simplify and solve the equation. Combine like terms, and then factor.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\left( x+4 \\right)\\left( x+4 \\right)=x+10\\\\{{x}^{2}}+8x+16=x+10\\\\{{x}^{2}}+8x-x+16-10=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{x}^{2}}+7x+6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left( x+6 \\right)\\left( x+1 \\right)=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSet each factor equal to zero and solve for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\left( x+6 \\right)=0\\,\\,\\text{or}\\,\\,\\left( x+1 \\right)=0\\\\x=-6\\text{ or }x=-1\\end{array}[\/latex]<\/p>\r\nNow check both solutions by substituting them into the original equation.\r\n\r\nSince [latex]x=\u22126[\/latex]\u00a0produces a false statement, it is an extraneous solution.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}-6+4=\\sqrt{-6+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=\\sqrt{4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=2\\\\\\text{FALSE!}\\\\\\\\\\\\-1+4=\\sqrt{-1+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=\\sqrt{9}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=3\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=\u22121[\/latex] is the only solution\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex] 4+\\sqrt{x+2}=x[\/latex]\r\n\r\n[reveal-answer q=\"568479\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568479\"]Isolate the radical term.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{x+2}=x-4[\/latex]<\/p>\r\nSquare both sides to remove the term [latex]x+2[\/latex]\u00a0from the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{x+2} \\right)}^{2}}={{\\left( x-4 \\right)}^{2}}[\/latex]<\/p>\r\nNow simplify and solve the equation. Combine like terms, and then factor.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x+2={{x}^{2}}-8x+16\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-8x-x+16-2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-9x+14\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=\\left( x-7 \\right)\\left( x-2 \\right)\\end{array}[\/latex]<\/p>\r\nSet each factor equal to zero and solve for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\left( x-7 \\right)=0\\text{ or }\\left( x-2 \\right)=0\\\\x=7\\text{ or }x=2\\end{array}[\/latex][latex] [\/latex]<\/p>\r\nNow check both solutions by substituting them into the original equation.\r\n\r\nSince [latex]x=2[\/latex]\u00a0produces a false statement, it is an extraneous solution.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}4+\\sqrt{7+2}=7\\\\4+\\sqrt{9}=7\\\\4+3=7\\\\7=7\\\\\\text{TRUE!}\\\\\\\\4+\\sqrt{2+2}=2\\\\4+\\sqrt{4}=2\\\\4+2=2\\\\6=2\\\\\\text{FALSE!}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=7[\/latex]\u00a0is the only solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last video example we solve a radical equation with a binomial term on one side.\r\nhttps:\/\/youtu.be\/y2yHPfuL0Hs\r\n<h2>Square Roots and Completing the Square<\/h2>\r\nQuadratic equations can be solved in many ways. In the previous section we introduced the idea that solutions to radical equations in general can be found using these facts:\r\n<div class=\"textbox shaded\">\r\n<h4>Solutions to Quadratic\u00a0Equations<\/h4>\r\nThe solutions of [latex]x^2=a[\/latex] are called the square roots of a.\r\n<ul>\r\n \t<li>When a is positive, a &gt; 0, [latex]x^2=a[\/latex] has two solutions, [latex]+\\sqrt{a},-\\sqrt{a}[\/latex]. [latex]+\\sqrt{a}[\/latex] is the nonnegative square root of a, and [latex]-\\sqrt{a}[\/latex] is the negative square root of a.<\/li>\r\n \t<li>When a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions.<\/li>\r\n \t<li>When a is zero, a = 0, [latex]x^2=a[\/latex] has one solution: a = 0<\/li>\r\n<\/ul>\r\n<\/div>\r\nA shortcut way to write [latex] \\sqrt{a}[\/latex] or [latex] -\\sqrt{a}[\/latex] is [latex] \\pm \\sqrt{a}[\/latex]. The symbol [latex]\\pm[\/latex] is often read \u201cpositive or negative.\u201d If it is used as an operation (addition or subtraction), it is read \u201cplus or minus.\u201d\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve using the Square Root Property. [latex]x^{2}=9[\/latex]\r\n\r\n[reveal-answer q=\"793132\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"793132\"]Since one side is simply [latex]x^{2}[\/latex], you can take the square root of both sides to get <em>x<\/em> on one side. Don\u2019t forget to use both positive and negative square roots!\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}=9\\\\\\,\\,\\,x=\\pm\\sqrt{9}\\\\\\,\\,\\,x=\\pm3\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=\\pm3[\/latex] (that is, [latex]x=3[\/latex] or [latex]-3[\/latex])\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that there is a difference here in solving [latex]x^{2}=9[\/latex]\u00a0and finding [latex] \\sqrt{9}[\/latex]. For [latex]x^{2}=9[\/latex], you are looking for <i>all numbers <\/i>whose square is 9. For [latex] \\sqrt{9}[\/latex], you only want the <i>principal<\/i> (nonnegative) square root. The negative of the principal square root is [latex] -\\sqrt{9}[\/latex]; both would be [latex] \\pm \\sqrt{9}[\/latex]. <i>Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!<\/i>\r\n\r\nIn the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating [latex]x^{2}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex]10x^{2}+5=85[\/latex]\r\n\r\n[reveal-answer q=\"637209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637209\"]If you try taking the square root of both sides of the original equation, you will have [latex] \\sqrt{10{{x}^{2}}+5}[\/latex] on the left, and you can\u2019t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[\/latex]\u00a0term by itself.\r\n<p style=\"text-align: center;\">[latex]10x^{2}+5=85[\/latex]<\/p>\r\nYou could now take the square root of both sides, but you would have [latex] \\sqrt{10}[\/latex]\u00a0as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.\r\n<p style=\"text-align: center;\">[latex]10x^{2}=80[\/latex]<\/p>\r\nNow you have only [latex]x^{2}[\/latex]\u00a0on the left, so you can use the Square Root Property easily.\r\n\r\nBe sure to simplify the radical if possible.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}{{x}^{2}}=8\\\\\\,\\,\\,x=\\pm \\sqrt{8}\\\\\\,\\,\\,\\,\\,\\,=\\pm \\sqrt{(4)(2)}\\\\\\,\\,\\,\\,\\,\\,=\\pm \\sqrt{4}\\sqrt{2}\\\\\\,\\,\\,\\,\\,\\,=\\pm 2\\sqrt{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=\\pm 2\\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of solving simple quadratic equations using square roots.\r\n\r\nhttps:\/\/youtu.be\/Fj-BP7uaWrI\r\n\r\nSometimes more than just the <i>x<\/i> is being squared:\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve.\u00a0[latex]\\left(x\u20132\\right)^{2}\u201350=0[\/latex]\r\n\r\n[reveal-answer q=\"347487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"347487\"]Again, taking the square root of both sides at this stage will leave something you can\u2019t work with on the left. Start by adding 50 to both sides.\r\n<p style=\"text-align: center;\">[latex]\\left(x-2\\right)^{2}-50=0[\/latex]<\/p>\r\nBecause [latex]\\left(x\u20132\\right)^{2}[\/latex] is a squared quantity, you can take the square root of both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left(x-2\\right)^{2}=50\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\x-2=\\pm\\sqrt{50}\\end{array}[\/latex]<\/p>\r\nTo isolate <i>x<\/i> on the left, you need to add 2 to both sides.\r\n\r\nBe sure to simplify the radical if possible.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x=2\\pm \\sqrt{50}\\\\\\,\\,\\,\\,=2\\pm \\sqrt{(25)(2)}\\\\\\,\\,\\,\\,=2\\pm \\sqrt{25}\\sqrt{2}\\\\\\,\\,\\,\\,=2\\pm 5\\sqrt{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=2\\pm 5\\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this video example, you will see more examples of solving quadratic equations using square roots.\r\n\r\nhttps:\/\/youtu.be\/4H5qZ_-8YM4\r\n<h3>Completing the Square to Solve a Quadratic Equation<\/h3>\r\nOf course, quadratic equations often will not come in the format of the examples above. Most of them will have <i>x<\/i> terms. However, you may be able to factor the expression into a squared binomial\u2014and if not, you can still use squared binomials to help you.\r\n\r\nSome of the above examples have squared binomials: [latex]\\left(1+r\\right)^{2}[\/latex]\u00a0and [latex]\\left(x\u20132\\right)^{2}[\/latex]\u00a0are squared binomials. If you expand these, you get a perfect square trinomial.\r\n\r\nPerfect square trinomials have the form [latex]x^{2}+2xs+s^{2}[\/latex]\u00a0and can be factored as [latex]\\left(x+s\\right)^{2}[\/latex], or they have the form [latex]x^{2}\u20132xs+s^{2}[\/latex] and can be factored as [latex]\\left(x\u2013s\\right)^{2}[\/latex]. Let\u2019s factor a perfect square trinomial into a squared binomial.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]9x^{2}\u201324x+16[\/latex].\r\n\r\n[reveal-answer q=\"844629\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"844629\"]First notice that the [latex]x^{2}[\/latex]\u00a0term and the constant term are both perfect squares.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}9x^{2}=\\left(3x\\right)^{2}\\\\\\,\\,\\,16=4^{2}\\end{array}[\/latex]<\/p>\r\nThen notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.\r\n<p style=\"text-align: center;\">[latex]24x=2\\left(3x\\right)\\left(4\\right)[\/latex]<\/p>\r\nA trinomial in the form [latex]r^{2}-2rs+s^{2}[\/latex]\u00a0can be factored as\u00a0[latex](r\u2013s)^{2}[\/latex].\r\n\r\nIn this case, the middle term is subtracted, so subtract <i>r<\/i> and <i>s<\/i> and square it to get\u00a0[latex](r\u2013s)^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\,\\,\\,r=3x\\\\s=4\\\\9x^{2}-24x+16=\\left(3x-4\\right)^{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(3x\u20134\\right)^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can use the procedure in this next example to help you solve equations where you identify perfect square trinomials, even if the equation is not set equal to 0.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. [latex]4x^{2}+20x+25=8[\/latex]\r\n\r\n[reveal-answer q=\"538757\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538757\"]Since there\u2019s an <i>x<\/i> term, you can\u2019t use the Square Root Property immediately (or even after adding or dividing by a constant).\r\n\r\nNotice, however, that the [latex]x^{2}[\/latex]\u00a0and constant terms on the left are both perfect squares: [latex]\\left(2x\\right)^{2}[\/latex]\u00a0and [latex]5^{2}[\/latex]. Check the middle term: is it [latex]2\\left(2x\\right)\\left(5\\right)[\/latex]? Yes!\r\n<p style=\"text-align: center;\">[latex]4x^{2}+20x+25=8[\/latex]<\/p>\r\nA trinomial in the form [latex]r^{2}+2rs+s^{2}[\/latex] can be factored as [latex]\\left(r+s\\right)^{2}[\/latex], so rewrite the left side as a squared binomial.\r\n<p style=\"text-align: center;\">[latex](2x+5)^{2}=8[\/latex]<\/p>\r\nNow you <i>can<\/i> use the Square Root Property. Some additional steps are needed to isolate <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}2x+5=\\pm \\sqrt{8}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\2x=-5\\pm \\sqrt{8}\\,\\,\\,\\,\\,\\\\\\\\x=-\\frac{5}{2}\\pm \\frac{1}{2}\\sqrt{8}\\end{array}[\/latex]<\/p>\r\nSimplify the radical when possible.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x=-\\frac{5}{2}\\pm \\frac{1}{2}\\sqrt{4}\\sqrt{2}\\\\\\\\x=-\\frac{5}{2}\\pm \\frac{1}{2}(2)\\sqrt{2}\\\\\\\\x=-\\frac{5}{2}\\pm \\sqrt{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=-\\frac{5}{2}\\pm \\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOne way to solve quadratic equations is by <strong>completing the square<\/strong>. When you don\u2019t have a perfect square trinomial, you can <i>create<\/i> one by adding a constant term that is a perfect square to both sides of the equation. Let\u2019s see how to find that constant term.\r\n\r\n\u201cCompleting the square\u201d does exactly what it says\u2014it takes something that is not a square and makes it one. This idea can be illustrated using an area model of the binomial [latex]x^{2}+bx[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064550\/image042-2.gif\" alt=\"X times X is x squared. X times b is bx. x(x+b)=x^{2}+bx\" width=\"321\" height=\"59\" \/>\r\n\r\nIn this example, the area of the overall rectangle is given by [latex]x\\left(x+b\\right)[\/latex].\r\n\r\nNow let's make this rectangle into a square. First, divide the red rectangle with area <i>bx<\/i> into two equal rectangles each with area [latex] \\frac{b}{2}x[\/latex]. Then rotate and reposition one of them. You haven't changed the size of the red area\u2014it still adds up to\u00a0[latex]bx[\/latex].\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064551\/image043-1.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x.\" width=\"282\" height=\"97\" \/><\/td>\r\n<td><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064554\/image044-1.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x. B\\2 times b\\2 is b\\2 squared. X squared plus 2 times b\\2 x + b\\2 squared\" width=\"279\" height=\"156\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe red rectangles now make up two sides of a square, shown in white. The area of that square is the length of the red rectangles squared, or [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex].\r\n\r\nHere comes the cool part\u2014do you see that when the white square is added to the blue and red regions, the whole shape is also now a square? In other words, you've \"completed the square!\" By adding the quantity\u00a0[latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex] to the original binomial, you've made a square, a square with sides of [latex]x+\\frac{b}{2}[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064556\/image047-2.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x. B\\2 times b\\2 is b\\2 squared. The lengths of the overall square is now x + b\\2. X squared +bx+ b\\2 squared equals x squared + 2 times b\\2 x + b\\2 squared = the square of x + b\\2.\" width=\"522\" height=\"206\" \/>\r\n\r\nNotice that the area of this square can be written as a squared binomial: [latex]\\left(x+\\frac{b}{2}\\right)^{2}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Finding a Value that will Complete the Square in an Expression<\/h3>\r\nTo complete the square for an expression of the form [latex]x^{2}+bx[\/latex]:\r\n<ul type=\"disc\">\r\n \t<li>Identify the value of <i>b;<\/i><\/li>\r\n \t<li>Calculate and add [latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex].<\/li>\r\n<\/ul>\r\nThe expression becomes [latex]x^{2}+bx+\\left(\\frac{b}{2}\\right)^{2}=\\left(x+\\frac{b}{2}\\right)^{2}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the number to add to [latex]x^{2}+8x[\/latex]\u00a0to make it a perfect square trinomial.\r\n\r\n[reveal-answer q=\"691356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"691356\"]First identify <i>b<\/i> if this has the form [latex]x^{2}+bx[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+8x\\\\b=8\\end{array}[\/latex]<\/p>\r\nTo complete the square, add [latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]b=8[\/latex], so [latex]\\left(\\frac{b}{2}\\right)^{2}=\\left(\\frac{8}{2}\\right)^{2}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+8x+\\left(4\\right)^{2}\\\\x^{2}+8x+16\\end{array}[\/latex]<\/p>\r\nCheck that the result is a perfect square trinomial. [latex]\\left(x+4\\right)^{2}=x^{2}+4x+4x+16=x^{2}+8x+16[\/latex], so it is.\r\n<h4>Answer<\/h4>\r\nAdding [latex]+16[\/latex] will make [latex]x^{2}+8x[\/latex]\u00a0a perfect square trinomial.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] is always positive, since it is the square of a number. When you complete the square, you are always adding a positive value.\r\n\r\nIn the following video, we show more examples of how to find a constant terms that will make a trinomial a perfect square.\r\n\r\nhttps:\/\/youtu.be\/vt-pM1LEP1M\r\n\r\nYou can use completing the square to help you solve a quadratic equation that cannot be solved by factoring.\r\n\r\nLet\u2019s start by seeing what happens when you complete the square in an equation. In the example below, notice that completing the square will result in adding a number to <i>both<\/i> sides of the equation\u2014you have to do this in order to keep both sides equal!\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRewrite [latex]x^{2}+6x=8[\/latex]\u00a0so that the left side is a perfect square trinomial.\r\n\r\n[reveal-answer q=\"539170\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"539170\"]This equation has a constant of 8. Ignore it for now and focus on the [latex]x^{2}[\/latex]\u00a0and <i>x<\/i> terms on the left side of the equation. The left side has the form [latex]x^{2}+bx[\/latex], so you can identify <i>b<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+6x=8\\\\b=6\\end{array}[\/latex]<\/p>\r\nTo complete the square, add [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] to the left side.\r\n\r\n[latex]b=6[\/latex], so [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}={{\\left( \\frac{6}{2} \\right)}^{2}}={{3}^{2}}=9.[\/latex]\r\n\r\nThis is an equation, though, so you must add the same number to the <i>right<\/i> side as well.\r\n<p style=\"text-align: center;\">[latex]x^{2}+6x+9=8+9[\/latex]<\/p>\r\nSimplify.\u00a0Check that the left side is a perfect square trinomial. [latex]\\begin{array}{r}\\left(x+3\\right)^{2}=x^{2}+3x+3x+9=x^{2}+6x+9\\end{array}{r}[\/latex], so it is.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+6x+9=17\\\\x^{2}+6x+9=17\\\\(x+3)^{2}=17\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x^{2}+6x+9=17[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nCan you see that completing the square in an equation is very similar to completing the square in an expression? The main difference is that you have to add the new number ([latex]+9[\/latex] in this case) to both sides of the equation to maintain equality.\r\n\r\nNow let\u2019s look at an example where you are using completing the square to actually solve an equation, finding a value for the variable.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve.\u00a0[latex]x^{2}\u201312x\u20134=0[\/latex]\r\n\r\n[reveal-answer q=\"903321\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"903321\"]Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.\r\n\r\nRewrite the equation with the left side in the form [latex]x^{2}+bx[\/latex]<i>,<\/i> to prepare to complete the square. Identify <i>b<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-12x=4\\,\\,\\,\\,\\,\\,\\,\\,\\\\b=-12\\end{array}[\/latex]<\/p>\r\nFigure out what value to add to complete the square. Add [latex] {{\\left( \\frac{b}{2}\\right)}^{2}}[\/latex] to complete the square, so [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}={{\\left( \\frac{-12}{2} \\right)}^{2}}={{\\left( -6 \\right)}^{2}}=36[\/latex].\r\n\r\nAdd the value to <i>both<\/i> sides of the equation and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-12x+36=4+36\\\\x^{2}-12x+36=40\\end{array}[\/latex]<\/p>\r\nRewrite the left side as a squared binomial.\r\n<p style=\"text-align: center;\">[latex]\\left(x-6\\right)^{2}=40[\/latex]<\/p>\r\nUse the Square Root Property. Remember to include both the positive and negative square root, or you\u2019ll miss one of the solutions.\r\n<p style=\"text-align: center;\">[latex] x-6=\\pm\\sqrt{40}[\/latex]<\/p>\r\nSolve for <i>x<\/i> by adding 6 to both sides. Simplify as needed.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x=6\\pm \\sqrt{40}\\\\\\,\\,\\,\\,=6\\pm \\sqrt{4}\\sqrt{10}\\\\\\,\\,\\,\\,=6\\pm 2\\sqrt{10}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x=6\\pm 2\\sqrt{10}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this last video, we solve more quadratic equations by completing the square.\r\n\r\nhttps:\/\/youtu.be\/IjCjbtrPWHM\r\n\r\nYou may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that\u2019s slightly different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]x^{2}+16x+17=-47[\/latex].\r\n\r\n[reveal-answer q=\"270245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"270245\"]Rewrite the equation so the left side has the form [latex]x^{2}+bx[\/latex]. Identify <i>b<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+16x=-64\\\\b=16\\end{array}[\/latex]<\/p>\r\nAdd [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex], which is [latex] {{\\left( \\frac{16}{2} \\right)}^{2}}={{8}^{2}}=64[\/latex], to both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+16x+64=-64+64\\\\x^{2}+16x+64=0\\end{array}[\/latex]<\/p>\r\nWrite the left side as a squared binomial.\r\n<p style=\"text-align: center;\">[latex]\\left(x+8\\right)^{2}=0[\/latex]<\/p>\r\nTake the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.\r\n<p style=\"text-align: center;\">[latex]x+8=0[\/latex]<\/p>\r\nSolve for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex]x=-8[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=-8[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTake a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation [latex]x^{2}+16x+17=\u221247[\/latex]\u00a0becomes [latex]x^{2}+16x+64=0[\/latex]. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).\r\n\r\nIt can be factored as [latex](x+8)(x+8)=0[\/latex], of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.\r\n<h2>Use the Quadratic Formula to Solve Quadratic Equations<\/h2>\r\n[caption id=\"attachment_5188\" align=\"aligncenter\" width=\"465\"]<img class=\" wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/24025157\/Screen-Shot-2016-06-23-at-7.51.28-PM-300x216.png\" alt=\"x is a fraction with the numerator -b plus or minus the square root of b squared minus 4 times a times c and the denominator is 2 times a\" width=\"465\" height=\"335\" \/> Quadratic formula[\/caption]\r\n\r\nYou can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for <i>x<\/i>, you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.\r\n\r\nThis formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].\r\n\r\nThe form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it's <i>vital<\/i> that you be sure the equation is in this form. If you don't, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRewrite the equation [latex]3x+2x^{2}+4=5[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.\r\n\r\n[reveal-answer q=\"489648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"489648\"]First be sure that the right side of the equation is 0. In this case, all you need to do is subtract 5 from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}3x+2x^{2}+4=5\\\\3x+2x^{2}+4\u20135=5\u20135\\end{array}[\/latex]<\/p>\r\nSimplify, and write the terms with the exponent on the variable in descending order.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+2x^{2}-1=0\\\\2x^{2}+3x-1=0\\end{array}[\/latex]<\/p>\r\nNow that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant. Note that since the constant 1 is subtracted, <i>c <\/i>must be negative.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,3x\\,\\,\\,-\\,\\,\\,1\\,\\,\\,=\\,\\,\\,0\\\\\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,ax^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,bx\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,a=2,\\,\\,b=3,\\,\\,c=-1\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]2x^{2}+3x\u20131=0;a=2,b=3,c=\u22121[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRewrite the equation [latex]2(x+3)^{2}\u20135x=6[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.\r\n\r\n[reveal-answer q=\"585220\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"585220\"]First be sure that the right side of the equation is 0.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2\\left(x+3\\right)^{2}\u20135x=6\\\\2(x+3)^{2}\u20135x\u20136=6\u20136\\end{array}[\/latex]<\/p>\r\nExpand the squared binomial, then simplify by combining like terms.\r\n\r\nBe sure to write the terms with the exponent on the variable in descending order.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(x^{2}+6x+9\\right)-5x-6=0\\\\2x^{2}+12x+18\u20135x\u20136=0\\\\2x^{2}+12x\u20135x+18\u20136=0\\\\2x^{2}+7x+12=0\\end{array}[\/latex]<\/p>\r\nNow that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,7x\\,\\,\\,+\\,\\,\\,12\\,\\,\\,=\\,\\,\\,0\\\\\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,\\,a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,\\,\\,\\,\\,a=2,\\,\\,b=7,\\,\\,c=7\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]2x^{2}+7x+12=0;\\,\\,a=2,b=7,c=12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\r\nThe Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.\r\n<ul>\r\n \t<li>Put the equation in standard form first.<\/li>\r\n \t<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be careful to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\r\n \t<li>Substitute the values for the coefficients into the Quadratic Formula.<\/li>\r\n \t<li>Simplify as much as possible.<\/li>\r\n \t<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted<i>.<\/i><\/li>\r\n \t<li>Simplify both values to get the possible solutions.<\/li>\r\n<\/ul>\r\nThat's a lot of steps. Let\u2019s try using the Quadratic Formula to solve a relatively simple equation first; then you\u2019ll go back and solve it again using another factoring method.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].\r\n\r\n[reveal-answer q=\"296770\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296770\"]First write the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1,b=4,c=-5\\end{array}[\/latex]<\/p>\r\nNote that the subtraction sign means the constant <i>c<\/i> is negative.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\u00a0[latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify, being careful to get the signs correct.\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\r\nSimplify some more.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\r\nSimplify the radical: [latex] \\sqrt{36}=6[\/latex].\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\r\nSeparate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nYou get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You\u2019ve solved the equation successfully using the Quadratic Formula!\r\n\r\nThe power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.\r\n\r\nIn teh following video, we show an example of using the quadratic formula to solve an equation with two real solutions.\r\n\r\nhttps:\/\/youtu.be\/xtwO-n8lRPw\r\n\r\nMost of the quadratic equations you've looked at have two solutions, like the one above. The following example is a little different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].\r\n\r\n[reveal-answer q=\"998241\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"998241\"]Subtract 6<i>x <\/i>from each side and add 16 to both sides to put the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\r\nIdentify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\r\nSince the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAgain, check using the original equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nIn the following video we show an example of using the quadratic formula to solve a quadratic\u00a0equation that has one repeated solution.\r\n\r\nhttps:\/\/youtu.be\/OXwwzWcxFgE\r\n\r\nIn this video example we show that solutions to quadratic equations can have rational answers.\r\n\r\nhttps:\/\/youtu.be\/xtwO-n8lRPw\r\n<h2>Applying the Quadratic Formula<\/h2>\r\nQuadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.\r\n\r\nA very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation isn't completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. (The negative value means it's heading toward the ground.)\r\n\r\nThe equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after <i>t<\/i> seconds. About how long does it take for the ball to hit the ground?\r\n\r\n[reveal-answer q=\"704677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704677\"]When the ball hits the ground, the height is 0. Substitute 0 for <i>h<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\r\nThis equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is <i>t<\/i> rather than <i>x<\/i>. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].\r\n<p style=\"text-align: center;\">[latex] t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\r\nSimplify. Be very careful with the signs.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\r\nUse a calculator to find both roots.\r\n<p style=\"text-align: center;\"><i>t<\/i> is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\r\nConsider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.\r\n<h4>Answer<\/h4>\r\nThe ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nBob made a quilt that is 4 ft [latex]\\times[\/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)\r\n\r\n[reveal-answer q=\"932211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932211\"]Sketch the problem. Since you don\u2019t know the width of the border, you will let the variable <i>x <\/i>represent the width.\r\n\r\nIn the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/>\r\n\r\nSince each side of the original 4 by 5 quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].\r\n\r\n(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/>\r\n\r\nYou are only interested in the area of the border strips. Write an expression for the area of the border.\r\n<p style=\"text-align: center;\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\r\n<p style=\"text-align: center;\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\r\nThere are 10 sq ft of fabric for the border, so set the area of border to be 10.\r\n<p style=\"text-align: center;\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\r\nMultiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]10=18x+4x^{2}[\/latex]<\/p>\r\nSubtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}-10\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\r\nFactor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\r\nUse the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\r\nFind the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\r\nIgnore the solution [latex]x=\u22125[\/latex], since the width could not be negative.\r\n<h4>Answer<\/h4>\r\nThe width of the border should be 0.5 ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this last video, we show how to use the quadratic formula to solve an application involving a picture frame.\r\n\r\nhttps:\/\/youtu.be\/AlIoxXQ-V50\r\n<h2>Summary<\/h2>\r\nA common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful\u2014when both sides of an equation are raised to an <i>even<\/i> power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.\r\n\r\nCompleting the square is used to change a binomial of the form [latex]x^{2}+bx[\/latex] into a perfect square trinomial [latex] {{x}^{2}}+bx+{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex], which can be factored to [latex] {{\\left( x+\\frac{b}{2} \\right)}^{2}}[\/latex]. When solving quadratic equations by completing the square, be careful to add [latex] {{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] to <i>both<\/i> sides of the equation to maintain equality. The Square Root Property can then be used to solve for <i>x<\/i>. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.\r\n\r\nQuadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations, or any other quadratic equation! The Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex] [\/latex].\r\n<h2><\/h2>","rendered":"<p>&nbsp;<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve Radical Equations\n<ul>\n<li>Isolate square roots in equations and solve for a variable<\/li>\n<li>Identify extraneous solutions to radical equations<\/li>\n<\/ul>\n<\/li>\n<li>Square Roots and Completing the Square for Solving Radical Equations\n<ul>\n<li>Use square roots to solve quadratic\u00a0equations<\/li>\n<li>Complete the square to solve a quadratic equation<\/li>\n<\/ul>\n<\/li>\n<li>Using the Quadratic Formula to Solve Quadratic Equations\n<ul>\n<li>Write a quadratic equation in standard form and identify the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> in a\u00a0standard form quadratic equation.<\/li>\n<li>Use the Quadratic Formula to find all real solutions.<\/li>\n<li>Solve application problems requiring the use of the Quadratic Formula.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations\u2014rearrange the expression to isolate the variable you want to know, and then solve the resulting equation.<\/p>\n<div class=\"textbox shaded\">\n<h4>Solutions to Radical Equations<\/h4>\n<p>The solutions of [latex]x^2=a[\/latex] are called the square roots of a.<\/p>\n<ul>\n<li>When a is positive, a &gt; 0, [latex]x^2=a[\/latex] has two solutions, [latex]+\\sqrt{a},-\\sqrt{a}[\/latex]. [latex]+\\sqrt{a}[\/latex] is the nonnegative square root of a, and [latex]-\\sqrt{a}[\/latex] is the negative square root of a.<\/li>\n<li>When a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions.<\/li>\n<li>When a is zero, a = 0, [latex]x^2=a[\/latex] has one solution: a = 0<\/li>\n<\/ul>\n<\/div>\n<p>Just to drive home the importance of the concept that when a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions, we will restate it in words. \u00a0If you have a negative number under a square root sign as in this example,<\/p>\n<p>[latex]\\sqrt{-3}[\/latex]<\/p>\n<p>There will be no real number solutions.<\/p>\n<p>There are two key ideas that you will be using to solve radical equations. The first is that if [latex]a=b[\/latex], then [latex]{{a}^{2}}={{b}^{2}}[\/latex]. (This property allows you to square both sides of an equation and remain certain that the two sides are still equal.) The second is that if the square root of any nonnegative number <i>x<\/i> is squared, then you get <i>x<\/i>: [latex]{{\\left( \\sqrt{x} \\right)}^{2}}=x[\/latex]. (This property allows you to \u201cremove\u201d the radicals from your equations.)<\/p>\n<p>Let\u2019s start with a radical equation that you can solve in a few steps:[latex]\\sqrt{x}-3=5[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]\\sqrt{x}-3=5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q946356\">Show Solution<\/span><\/p>\n<div id=\"q946356\" class=\"hidden-answer\" style=\"display: none\">Add 3 to both sides to isolate the variable term on the left side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\n<p>Collect like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=8[\/latex]<\/p>\n<p>Square both sides to remove the radical, since [latex]{{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the 8 also! Then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=64[\/latex] is the solution to [latex]\\sqrt{x}-3=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>To check your solution, you can substitute 64 in for <i>x<\/i> in the original equation. Does [latex]\\sqrt{64}-3=5[\/latex]? Yes\u2014the square root of 64 is 8, and [latex]8\u22123=5[\/latex].<\/p>\n<p>Notice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.<\/p>\n<p>In the example above, only the variable <i>x<\/i> was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>. Watch how the next two problems are solved.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]\\sqrt{x+8}=3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q673245\">Show Solution<\/span><\/p>\n<div id=\"q673245\" class=\"hidden-answer\" style=\"display: none\">Notice how the radical contains a binomial: [latex]x+8[\/latex]. Square both sides to remove the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{x+8} \\right)}^{2}}={{\\left( 3 \\right)}^{2}}[\/latex]<\/p>\n<p>[latex]{{\\left( \\sqrt{x+8} \\right)}^{2}}=x+8[\/latex]. Now simplify the equation and solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+8=9\\\\x=1\\end{array}[\/latex]<\/p>\n<p>Check your answer. Substituting 1 for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{1+8}=3\\\\\\sqrt{9}=3\\\\3=3\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=1[\/latex] is the solution to [latex]\\sqrt{x+8}=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show how to solve simple radical equations.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Solve a Basic Radical Equation - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tT0Zwsto6AQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]1+\\sqrt{2x+3}=6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479262\">Show Solution<\/span><\/p>\n<div id=\"q479262\" class=\"hidden-answer\" style=\"display: none\">Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Simplify the equation and solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\n<p>Check your answer. Substituting 11 for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=11[\/latex] is the solution for [latex]1+\\sqrt{2x+3}=6[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solving Radical Equations<\/h3>\n<p>Follow the following four steps to solve radical equations.<\/p>\n<ol>\n<li>Isolate the radical expression.<\/li>\n<li>Square both sides of the equation: If [latex]x=y[\/latex] then [latex]x^{2}=y^{2}[\/latex].<\/li>\n<li>Once the radical is removed, solve for the unknown.<\/li>\n<li>Check all answers.<\/li>\n<\/ol>\n<\/div>\n<h2>Identify Extraneous\u00a0Solutions<\/h2>\n<p>Following rules is important, but so is paying attention to the math in front of you\u2014especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]\\sqrt{a-5}=-2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798652\">Show Solution<\/span><\/p>\n<div id=\"q798652\" class=\"hidden-answer\" style=\"display: none\">Square both sides to remove the term [latex]a\u20135[\/latex] from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{a-5} \\right)}^{2}}={{(-2)}^{2}}[\/latex]<\/p>\n<p>Write the simplified equation, and solve for <i>a<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}a-5=4\\\\a=9\\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Now check the solution by substituting [latex]a=9[\/latex] into the original equation.<\/p>\n<p>It does not check!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{9-5}=-2\\\\\\sqrt{4}=-2\\\\2\\ne -2\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>No solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Look at that\u2014the answer [latex]a=9[\/latex] does not produce a true statement when substituted back into the original equation. What happened?<\/p>\n<p>Check the original problem: [latex]\\sqrt{a-5}=-2[\/latex]. Notice that the radical is set equal to [latex]\u22122[\/latex], and recall that the principal square root of a number can only be <i>positive<\/i>. This means that no value for <i>a<\/i> will result in a radical expression whose positive square root is [latex]\u22122[\/latex]! You might have noticed that right away and concluded that there were no solutions for <i>a<\/i>.<\/p>\n<p>Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called <strong>extraneous solutions<\/strong>. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important\u2014if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.<br \/>\nIn the next video example, we solve more radical equations that may have extraneous solutions.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 2:  Solve Radical Equations - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qkZHKK77grM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Have a look at the following problem. Notice how the original problem is [latex]x+4=\\sqrt{x+10}[\/latex], but after both sides are squared, it becomes [latex]{{x}^{2}}+8x+16=x+10[\/latex]. Squaring both sides may have introduced an extraneous solution.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]x+4=\\sqrt{x+10}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705028\">Show Solution<\/span><\/p>\n<div id=\"q705028\" class=\"hidden-answer\" style=\"display: none\">Square both sides to remove the term [latex]x+10[\/latex]\u00a0from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( x+4 \\right)}^{2}}={{\\left( \\sqrt{x+10} \\right)}^{2}}[\/latex]<\/p>\n<p>Now simplify and solve the equation. Combine like terms, and then factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left( x+4 \\right)\\left( x+4 \\right)=x+10\\\\{{x}^{2}}+8x+16=x+10\\\\{{x}^{2}}+8x-x+16-10=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{x}^{2}}+7x+6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left( x+6 \\right)\\left( x+1 \\right)=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Set each factor equal to zero and solve for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left( x+6 \\right)=0\\,\\,\\text{or}\\,\\,\\left( x+1 \\right)=0\\\\x=-6\\text{ or }x=-1\\end{array}[\/latex]<\/p>\n<p>Now check both solutions by substituting them into the original equation.<\/p>\n<p>Since [latex]x=\u22126[\/latex]\u00a0produces a false statement, it is an extraneous solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-6+4=\\sqrt{-6+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=\\sqrt{4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=2\\\\\\text{FALSE!}\\\\\\\\\\\\-1+4=\\sqrt{-1+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=\\sqrt{9}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=3\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\u22121[\/latex] is the only solution<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]4+\\sqrt{x+2}=x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568479\">Show Solution<\/span><\/p>\n<div id=\"q568479\" class=\"hidden-answer\" style=\"display: none\">Isolate the radical term.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x+2}=x-4[\/latex]<\/p>\n<p>Square both sides to remove the term [latex]x+2[\/latex]\u00a0from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{x+2} \\right)}^{2}}={{\\left( x-4 \\right)}^{2}}[\/latex]<\/p>\n<p>Now simplify and solve the equation. Combine like terms, and then factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2={{x}^{2}}-8x+16\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-8x-x+16-2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-9x+14\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=\\left( x-7 \\right)\\left( x-2 \\right)\\end{array}[\/latex]<\/p>\n<p>Set each factor equal to zero and solve for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left( x-7 \\right)=0\\text{ or }\\left( x-2 \\right)=0\\\\x=7\\text{ or }x=2\\end{array}[\/latex][latex][\/latex]<\/p>\n<p>Now check both solutions by substituting them into the original equation.<\/p>\n<p>Since [latex]x=2[\/latex]\u00a0produces a false statement, it is an extraneous solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4+\\sqrt{7+2}=7\\\\4+\\sqrt{9}=7\\\\4+3=7\\\\7=7\\\\\\text{TRUE!}\\\\\\\\4+\\sqrt{2+2}=2\\\\4+\\sqrt{4}=2\\\\4+2=2\\\\6=2\\\\\\text{FALSE!}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=7[\/latex]\u00a0is the only solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the last video example we solve a radical equation with a binomial term on one side.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 4:  Solve Radical Equations - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/y2yHPfuL0Hs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Square Roots and Completing the Square<\/h2>\n<p>Quadratic equations can be solved in many ways. In the previous section we introduced the idea that solutions to radical equations in general can be found using these facts:<\/p>\n<div class=\"textbox shaded\">\n<h4>Solutions to Quadratic\u00a0Equations<\/h4>\n<p>The solutions of [latex]x^2=a[\/latex] are called the square roots of a.<\/p>\n<ul>\n<li>When a is positive, a &gt; 0, [latex]x^2=a[\/latex] has two solutions, [latex]+\\sqrt{a},-\\sqrt{a}[\/latex]. [latex]+\\sqrt{a}[\/latex] is the nonnegative square root of a, and [latex]-\\sqrt{a}[\/latex] is the negative square root of a.<\/li>\n<li>When a is negative, a &lt; 0, [latex]x^2=a[\/latex] has no solutions.<\/li>\n<li>When a is zero, a = 0, [latex]x^2=a[\/latex] has one solution: a = 0<\/li>\n<\/ul>\n<\/div>\n<p>A shortcut way to write [latex]\\sqrt{a}[\/latex] or [latex]-\\sqrt{a}[\/latex] is [latex]\\pm \\sqrt{a}[\/latex]. The symbol [latex]\\pm[\/latex] is often read \u201cpositive or negative.\u201d If it is used as an operation (addition or subtraction), it is read \u201cplus or minus.\u201d<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve using the Square Root Property. [latex]x^{2}=9[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q793132\">Show Solution<\/span><\/p>\n<div id=\"q793132\" class=\"hidden-answer\" style=\"display: none\">Since one side is simply [latex]x^{2}[\/latex], you can take the square root of both sides to get <em>x<\/em> on one side. Don\u2019t forget to use both positive and negative square roots!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}=9\\\\\\,\\,\\,x=\\pm\\sqrt{9}\\\\\\,\\,\\,x=\\pm3\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\\pm3[\/latex] (that is, [latex]x=3[\/latex] or [latex]-3[\/latex])<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that there is a difference here in solving [latex]x^{2}=9[\/latex]\u00a0and finding [latex]\\sqrt{9}[\/latex]. For [latex]x^{2}=9[\/latex], you are looking for <i>all numbers <\/i>whose square is 9. For [latex]\\sqrt{9}[\/latex], you only want the <i>principal<\/i> (nonnegative) square root. The negative of the principal square root is [latex]-\\sqrt{9}[\/latex]; both would be [latex]\\pm \\sqrt{9}[\/latex]. <i>Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!<\/i><\/p>\n<p>In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating [latex]x^{2}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]10x^{2}+5=85[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q637209\">Show Solution<\/span><\/p>\n<div id=\"q637209\" class=\"hidden-answer\" style=\"display: none\">If you try taking the square root of both sides of the original equation, you will have [latex]\\sqrt{10{{x}^{2}}+5}[\/latex] on the left, and you can\u2019t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[\/latex]\u00a0term by itself.<\/p>\n<p style=\"text-align: center;\">[latex]10x^{2}+5=85[\/latex]<\/p>\n<p>You could now take the square root of both sides, but you would have [latex]\\sqrt{10}[\/latex]\u00a0as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.<\/p>\n<p style=\"text-align: center;\">[latex]10x^{2}=80[\/latex]<\/p>\n<p>Now you have only [latex]x^{2}[\/latex]\u00a0on the left, so you can use the Square Root Property easily.<\/p>\n<p>Be sure to simplify the radical if possible.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{{x}^{2}}=8\\\\\\,\\,\\,x=\\pm \\sqrt{8}\\\\\\,\\,\\,\\,\\,\\,=\\pm \\sqrt{(4)(2)}\\\\\\,\\,\\,\\,\\,\\,=\\pm \\sqrt{4}\\sqrt{2}\\\\\\,\\,\\,\\,\\,\\,=\\pm 2\\sqrt{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\\pm 2\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of solving simple quadratic equations using square roots.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 1:  Solving Quadratic Equations Using Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Fj-BP7uaWrI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes more than just the <i>x<\/i> is being squared:<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve.\u00a0[latex]\\left(x\u20132\\right)^{2}\u201350=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q347487\">Show Solution<\/span><\/p>\n<div id=\"q347487\" class=\"hidden-answer\" style=\"display: none\">Again, taking the square root of both sides at this stage will leave something you can\u2019t work with on the left. Start by adding 50 to both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x-2\\right)^{2}-50=0[\/latex]<\/p>\n<p>Because [latex]\\left(x\u20132\\right)^{2}[\/latex] is a squared quantity, you can take the square root of both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left(x-2\\right)^{2}=50\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\x-2=\\pm\\sqrt{50}\\end{array}[\/latex]<\/p>\n<p>To isolate <i>x<\/i> on the left, you need to add 2 to both sides.<\/p>\n<p>Be sure to simplify the radical if possible.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=2\\pm \\sqrt{50}\\\\\\,\\,\\,\\,=2\\pm \\sqrt{(25)(2)}\\\\\\,\\,\\,\\,=2\\pm \\sqrt{25}\\sqrt{2}\\\\\\,\\,\\,\\,=2\\pm 5\\sqrt{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=2\\pm 5\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this video example, you will see more examples of solving quadratic equations using square roots.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 2:  Solving Quadratic Equations Using Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4H5qZ_-8YM4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Completing the Square to Solve a Quadratic Equation<\/h3>\n<p>Of course, quadratic equations often will not come in the format of the examples above. Most of them will have <i>x<\/i> terms. However, you may be able to factor the expression into a squared binomial\u2014and if not, you can still use squared binomials to help you.<\/p>\n<p>Some of the above examples have squared binomials: [latex]\\left(1+r\\right)^{2}[\/latex]\u00a0and [latex]\\left(x\u20132\\right)^{2}[\/latex]\u00a0are squared binomials. If you expand these, you get a perfect square trinomial.<\/p>\n<p>Perfect square trinomials have the form [latex]x^{2}+2xs+s^{2}[\/latex]\u00a0and can be factored as [latex]\\left(x+s\\right)^{2}[\/latex], or they have the form [latex]x^{2}\u20132xs+s^{2}[\/latex] and can be factored as [latex]\\left(x\u2013s\\right)^{2}[\/latex]. Let\u2019s factor a perfect square trinomial into a squared binomial.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]9x^{2}\u201324x+16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q844629\">Show Solution<\/span><\/p>\n<div id=\"q844629\" class=\"hidden-answer\" style=\"display: none\">First notice that the [latex]x^{2}[\/latex]\u00a0term and the constant term are both perfect squares.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}9x^{2}=\\left(3x\\right)^{2}\\\\\\,\\,\\,16=4^{2}\\end{array}[\/latex]<\/p>\n<p>Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.<\/p>\n<p style=\"text-align: center;\">[latex]24x=2\\left(3x\\right)\\left(4\\right)[\/latex]<\/p>\n<p>A trinomial in the form [latex]r^{2}-2rs+s^{2}[\/latex]\u00a0can be factored as\u00a0[latex](r\u2013s)^{2}[\/latex].<\/p>\n<p>In this case, the middle term is subtracted, so subtract <i>r<\/i> and <i>s<\/i> and square it to get\u00a0[latex](r\u2013s)^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\,\\,\\,r=3x\\\\s=4\\\\9x^{2}-24x+16=\\left(3x-4\\right)^{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(3x\u20134\\right)^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can use the procedure in this next example to help you solve equations where you identify perfect square trinomials, even if the equation is not set equal to 0.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. [latex]4x^{2}+20x+25=8[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538757\">Show Solution<\/span><\/p>\n<div id=\"q538757\" class=\"hidden-answer\" style=\"display: none\">Since there\u2019s an <i>x<\/i> term, you can\u2019t use the Square Root Property immediately (or even after adding or dividing by a constant).<\/p>\n<p>Notice, however, that the [latex]x^{2}[\/latex]\u00a0and constant terms on the left are both perfect squares: [latex]\\left(2x\\right)^{2}[\/latex]\u00a0and [latex]5^{2}[\/latex]. Check the middle term: is it [latex]2\\left(2x\\right)\\left(5\\right)[\/latex]? Yes!<\/p>\n<p style=\"text-align: center;\">[latex]4x^{2}+20x+25=8[\/latex]<\/p>\n<p>A trinomial in the form [latex]r^{2}+2rs+s^{2}[\/latex] can be factored as [latex]\\left(r+s\\right)^{2}[\/latex], so rewrite the left side as a squared binomial.<\/p>\n<p style=\"text-align: center;\">[latex](2x+5)^{2}=8[\/latex]<\/p>\n<p>Now you <i>can<\/i> use the Square Root Property. Some additional steps are needed to isolate <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+5=\\pm \\sqrt{8}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\2x=-5\\pm \\sqrt{8}\\,\\,\\,\\,\\,\\\\\\\\x=-\\frac{5}{2}\\pm \\frac{1}{2}\\sqrt{8}\\end{array}[\/latex]<\/p>\n<p>Simplify the radical when possible.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=-\\frac{5}{2}\\pm \\frac{1}{2}\\sqrt{4}\\sqrt{2}\\\\\\\\x=-\\frac{5}{2}\\pm \\frac{1}{2}(2)\\sqrt{2}\\\\\\\\x=-\\frac{5}{2}\\pm \\sqrt{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-\\frac{5}{2}\\pm \\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>One way to solve quadratic equations is by <strong>completing the square<\/strong>. When you don\u2019t have a perfect square trinomial, you can <i>create<\/i> one by adding a constant term that is a perfect square to both sides of the equation. Let\u2019s see how to find that constant term.<\/p>\n<p>\u201cCompleting the square\u201d does exactly what it says\u2014it takes something that is not a square and makes it one. This idea can be illustrated using an area model of the binomial [latex]x^{2}+bx[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064550\/image042-2.gif\" alt=\"X times X is x squared. X times b is bx. x(x+b)=x^{2}+bx\" width=\"321\" height=\"59\" \/><\/p>\n<p>In this example, the area of the overall rectangle is given by [latex]x\\left(x+b\\right)[\/latex].<\/p>\n<p>Now let&#8217;s make this rectangle into a square. First, divide the red rectangle with area <i>bx<\/i> into two equal rectangles each with area [latex]\\frac{b}{2}x[\/latex]. Then rotate and reposition one of them. You haven&#8217;t changed the size of the red area\u2014it still adds up to\u00a0[latex]bx[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064551\/image043-1.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x.\" width=\"282\" height=\"97\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064554\/image044-1.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x. B\\2 times b\\2 is b\\2 squared. X squared plus 2 times b\\2 x + b\\2 squared\" width=\"279\" height=\"156\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The red rectangles now make up two sides of a square, shown in white. The area of that square is the length of the red rectangles squared, or [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex].<\/p>\n<p>Here comes the cool part\u2014do you see that when the white square is added to the blue and red regions, the whole shape is also now a square? In other words, you&#8217;ve &#8220;completed the square!&#8221; By adding the quantity\u00a0[latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex] to the original binomial, you&#8217;ve made a square, a square with sides of [latex]x+\\frac{b}{2}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064556\/image047-2.gif\" alt=\"X times X is X squared. X times b\/2 is equal to b\/2 x. X times b\/2 is equal to b\/2 x. B\\2 times b\\2 is b\\2 squared. The lengths of the overall square is now x + b\\2. X squared +bx+ b\\2 squared equals x squared + 2 times b\\2 x + b\\2 squared = the square of x + b\\2.\" width=\"522\" height=\"206\" \/><\/p>\n<p>Notice that the area of this square can be written as a squared binomial: [latex]\\left(x+\\frac{b}{2}\\right)^{2}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Finding a Value that will Complete the Square in an Expression<\/h3>\n<p>To complete the square for an expression of the form [latex]x^{2}+bx[\/latex]:<\/p>\n<ul type=\"disc\">\n<li>Identify the value of <i>b;<\/i><\/li>\n<li>Calculate and add [latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex].<\/li>\n<\/ul>\n<p>The expression becomes [latex]x^{2}+bx+\\left(\\frac{b}{2}\\right)^{2}=\\left(x+\\frac{b}{2}\\right)^{2}[\/latex].<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the number to add to [latex]x^{2}+8x[\/latex]\u00a0to make it a perfect square trinomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q691356\">Show Solution<\/span><\/p>\n<div id=\"q691356\" class=\"hidden-answer\" style=\"display: none\">First identify <i>b<\/i> if this has the form [latex]x^{2}+bx[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+8x\\\\b=8\\end{array}[\/latex]<\/p>\n<p>To complete the square, add [latex]\\left(\\frac{b}{2}\\right)^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]b=8[\/latex], so [latex]\\left(\\frac{b}{2}\\right)^{2}=\\left(\\frac{8}{2}\\right)^{2}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+8x+\\left(4\\right)^{2}\\\\x^{2}+8x+16\\end{array}[\/latex]<\/p>\n<p>Check that the result is a perfect square trinomial. [latex]\\left(x+4\\right)^{2}=x^{2}+4x+4x+16=x^{2}+8x+16[\/latex], so it is.<\/p>\n<h4>Answer<\/h4>\n<p>Adding [latex]+16[\/latex] will make [latex]x^{2}+8x[\/latex]\u00a0a perfect square trinomial.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] is always positive, since it is the square of a number. When you complete the square, you are always adding a positive value.<\/p>\n<p>In the following video, we show more examples of how to find a constant terms that will make a trinomial a perfect square.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex:  Creating a Perfect Square Quadratic Trinomial Expression\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vt-pM1LEP1M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can use completing the square to help you solve a quadratic equation that cannot be solved by factoring.<\/p>\n<p>Let\u2019s start by seeing what happens when you complete the square in an equation. In the example below, notice that completing the square will result in adding a number to <i>both<\/i> sides of the equation\u2014you have to do this in order to keep both sides equal!<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rewrite [latex]x^{2}+6x=8[\/latex]\u00a0so that the left side is a perfect square trinomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q539170\">Show Solution<\/span><\/p>\n<div id=\"q539170\" class=\"hidden-answer\" style=\"display: none\">This equation has a constant of 8. Ignore it for now and focus on the [latex]x^{2}[\/latex]\u00a0and <i>x<\/i> terms on the left side of the equation. The left side has the form [latex]x^{2}+bx[\/latex], so you can identify <i>b<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+6x=8\\\\b=6\\end{array}[\/latex]<\/p>\n<p>To complete the square, add [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] to the left side.<\/p>\n<p>[latex]b=6[\/latex], so [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}={{\\left( \\frac{6}{2} \\right)}^{2}}={{3}^{2}}=9.[\/latex]<\/p>\n<p>This is an equation, though, so you must add the same number to the <i>right<\/i> side as well.<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+6x+9=8+9[\/latex]<\/p>\n<p>Simplify.\u00a0Check that the left side is a perfect square trinomial. [latex]\\begin{array}{r}\\left(x+3\\right)^{2}=x^{2}+3x+3x+9=x^{2}+6x+9\\end{array}{r}[\/latex], so it is.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+6x+9=17\\\\x^{2}+6x+9=17\\\\(x+3)^{2}=17\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x^{2}+6x+9=17[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Can you see that completing the square in an equation is very similar to completing the square in an expression? The main difference is that you have to add the new number ([latex]+9[\/latex] in this case) to both sides of the equation to maintain equality.<\/p>\n<p>Now let\u2019s look at an example where you are using completing the square to actually solve an equation, finding a value for the variable.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve.\u00a0[latex]x^{2}\u201312x\u20134=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q903321\">Show Solution<\/span><\/p>\n<div id=\"q903321\" class=\"hidden-answer\" style=\"display: none\">Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.<\/p>\n<p>Rewrite the equation with the left side in the form [latex]x^{2}+bx[\/latex]<i>,<\/i> to prepare to complete the square. Identify <i>b<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-12x=4\\,\\,\\,\\,\\,\\,\\,\\,\\\\b=-12\\end{array}[\/latex]<\/p>\n<p>Figure out what value to add to complete the square. Add [latex]{{\\left( \\frac{b}{2}\\right)}^{2}}[\/latex] to complete the square, so [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}={{\\left( \\frac{-12}{2} \\right)}^{2}}={{\\left( -6 \\right)}^{2}}=36[\/latex].<\/p>\n<p>Add the value to <i>both<\/i> sides of the equation and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-12x+36=4+36\\\\x^{2}-12x+36=40\\end{array}[\/latex]<\/p>\n<p>Rewrite the left side as a squared binomial.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x-6\\right)^{2}=40[\/latex]<\/p>\n<p>Use the Square Root Property. Remember to include both the positive and negative square root, or you\u2019ll miss one of the solutions.<\/p>\n<p style=\"text-align: center;\">[latex]x-6=\\pm\\sqrt{40}[\/latex]<\/p>\n<p>Solve for <i>x<\/i> by adding 6 to both sides. Simplify as needed.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=6\\pm \\sqrt{40}\\\\\\,\\,\\,\\,=6\\pm \\sqrt{4}\\sqrt{10}\\\\\\,\\,\\,\\,=6\\pm 2\\sqrt{10}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=6\\pm 2\\sqrt{10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this last video, we solve more quadratic equations by completing the square.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex 2:  Completing the Square - Real Irrational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/IjCjbtrPWHM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that\u2019s slightly different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]x^{2}+16x+17=-47[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q270245\">Show Solution<\/span><\/p>\n<div id=\"q270245\" class=\"hidden-answer\" style=\"display: none\">Rewrite the equation so the left side has the form [latex]x^{2}+bx[\/latex]. Identify <i>b<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}+16x=-64\\\\b=16\\end{array}[\/latex]<\/p>\n<p>Add [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex], which is [latex]{{\\left( \\frac{16}{2} \\right)}^{2}}={{8}^{2}}=64[\/latex], to both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+16x+64=-64+64\\\\x^{2}+16x+64=0\\end{array}[\/latex]<\/p>\n<p>Write the left side as a squared binomial.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+8\\right)^{2}=0[\/latex]<\/p>\n<p>Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.<\/p>\n<p style=\"text-align: center;\">[latex]x+8=0[\/latex]<\/p>\n<p>Solve for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]x=-8[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-8[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation [latex]x^{2}+16x+17=\u221247[\/latex]\u00a0becomes [latex]x^{2}+16x+64=0[\/latex]. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).<\/p>\n<p>It can be factored as [latex](x+8)(x+8)=0[\/latex], of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.<\/p>\n<h2>Use the Quadratic Formula to Solve Quadratic Equations<\/h2>\n<div id=\"attachment_5188\" style=\"width: 475px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5188\" class=\"wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/24025157\/Screen-Shot-2016-06-23-at-7.51.28-PM-300x216.png\" alt=\"x is a fraction with the numerator -b plus or minus the square root of b squared minus 4 times a times c and the denominator is 2 times a\" width=\"465\" height=\"335\" \/><\/p>\n<p id=\"caption-attachment-5188\" class=\"wp-caption-text\">Quadratic formula<\/p>\n<\/div>\n<p>You can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for <i>x<\/i>, you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.<\/p>\n<p>This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].<\/p>\n<p>The form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it&#8217;s <i>vital<\/i> that you be sure the equation is in this form. If you don&#8217;t, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rewrite the equation [latex]3x+2x^{2}+4=5[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q489648\">Show Solution<\/span><\/p>\n<div id=\"q489648\" class=\"hidden-answer\" style=\"display: none\">First be sure that the right side of the equation is 0. In this case, all you need to do is subtract 5 from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}3x+2x^{2}+4=5\\\\3x+2x^{2}+4\u20135=5\u20135\\end{array}[\/latex]<\/p>\n<p>Simplify, and write the terms with the exponent on the variable in descending order.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+2x^{2}-1=0\\\\2x^{2}+3x-1=0\\end{array}[\/latex]<\/p>\n<p>Now that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant. Note that since the constant 1 is subtracted, <i>c <\/i>must be negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,3x\\,\\,\\,-\\,\\,\\,1\\,\\,\\,=\\,\\,\\,0\\\\\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,ax^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,bx\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,a=2,\\,\\,b=3,\\,\\,c=-1\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2x^{2}+3x\u20131=0;a=2,b=3,c=\u22121[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rewrite the equation [latex]2(x+3)^{2}\u20135x=6[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q585220\">Show Solution<\/span><\/p>\n<div id=\"q585220\" class=\"hidden-answer\" style=\"display: none\">First be sure that the right side of the equation is 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2\\left(x+3\\right)^{2}\u20135x=6\\\\2(x+3)^{2}\u20135x\u20136=6\u20136\\end{array}[\/latex]<\/p>\n<p>Expand the squared binomial, then simplify by combining like terms.<\/p>\n<p>Be sure to write the terms with the exponent on the variable in descending order.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(x^{2}+6x+9\\right)-5x-6=0\\\\2x^{2}+12x+18\u20135x\u20136=0\\\\2x^{2}+12x\u20135x+18\u20136=0\\\\2x^{2}+7x+12=0\\end{array}[\/latex]<\/p>\n<p>Now that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,7x\\,\\,\\,+\\,\\,\\,12\\,\\,\\,=\\,\\,\\,0\\\\\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,\\,a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,\\,\\,\\,\\,a=2,\\,\\,b=7,\\,\\,c=7\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2x^{2}+7x+12=0;\\,\\,a=2,b=7,c=12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\n<p>The Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.<\/p>\n<ul>\n<li>Put the equation in standard form first.<\/li>\n<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be careful to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\n<li>Substitute the values for the coefficients into the Quadratic Formula.<\/li>\n<li>Simplify as much as possible.<\/li>\n<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted<i>.<\/i><\/li>\n<li>Simplify both values to get the possible solutions.<\/li>\n<\/ul>\n<p>That&#8217;s a lot of steps. Let\u2019s try using the Quadratic Formula to solve a relatively simple equation first; then you\u2019ll go back and solve it again using another factoring method.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q296770\">Show Solution<\/span><\/p>\n<div id=\"q296770\" class=\"hidden-answer\" style=\"display: none\">First write the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1,b=4,c=-5\\end{array}[\/latex]<\/p>\n<p>Note that the subtraction sign means the constant <i>c<\/i> is negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.\u00a0[latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify, being careful to get the signs correct.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\n<p>Simplify some more.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\n<p>Simplify the radical: [latex]\\sqrt{36}=6[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\n<p>Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>You get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You\u2019ve solved the equation successfully using the Quadratic Formula!<\/p>\n<p>The power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.<\/p>\n<p>In teh following video, we show an example of using the quadratic formula to solve an equation with two real solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Ex:  Quadratic Formula - Two Real Rational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xtwO-n8lRPw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Most of the quadratic equations you&#8217;ve looked at have two solutions, like the one above. The following example is a little different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q998241\">Show Solution<\/span><\/p>\n<div id=\"q998241\" class=\"hidden-answer\" style=\"display: none\">Subtract 6<i>x <\/i>from each side and add 16 to both sides to put the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\n<p>Identify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\n<p>Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Again, check using the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>In the following video we show an example of using the quadratic formula to solve a quadratic\u00a0equation that has one repeated solution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Ex:  Quadratic Formula - One Real Rational Repeated Solution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/OXwwzWcxFgE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this video example we show that solutions to quadratic equations can have rational answers.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Ex:  Quadratic Formula - Two Real Rational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xtwO-n8lRPw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Applying the Quadratic Formula<\/h2>\n<p>Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.<\/p>\n<p>A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation isn&#8217;t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. (The negative value means it&#8217;s heading toward the ground.)<\/p>\n<p>The equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after <i>t<\/i> seconds. About how long does it take for the ball to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704677\">Show Solution<\/span><\/p>\n<div id=\"q704677\" class=\"hidden-answer\" style=\"display: none\">When the ball hits the ground, the height is 0. Substitute 0 for <i>h<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\n<p>This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is <i>t<\/i> rather than <i>x<\/i>. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\n<p>Simplify. Be very careful with the signs.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\n<p>Use a calculator to find both roots.<\/p>\n<p style=\"text-align: center;\"><i>t<\/i> is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\n<p>Consider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.<\/p>\n<h4>Answer<\/h4>\n<p>The ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Bob made a quilt that is 4 ft [latex]\\times[\/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932211\">Show Solution<\/span><\/p>\n<div id=\"q932211\" class=\"hidden-answer\" style=\"display: none\">Sketch the problem. Since you don\u2019t know the width of the border, you will let the variable <i>x <\/i>represent the width.<\/p>\n<p>In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/><\/p>\n<p>Since each side of the original 4 by 5 quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].<\/p>\n<p>(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/><\/p>\n<p>You are only interested in the area of the border strips. Write an expression for the area of the border.<\/p>\n<p style=\"text-align: center;\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\n<p style=\"text-align: center;\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\n<p>There are 10 sq ft of fabric for the border, so set the area of border to be 10.<\/p>\n<p style=\"text-align: center;\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\n<p>Multiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]10=18x+4x^{2}[\/latex]<\/p>\n<p>Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}-10\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\n<p>Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\n<p>Use the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\n<p>Find the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\n<p>Ignore the solution [latex]x=\u22125[\/latex], since the width could not be negative.<\/p>\n<h4>Answer<\/h4>\n<p>The width of the border should be 0.5 ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this last video, we show how to use the quadratic formula to solve an application involving a picture frame.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Ex: Quadratic Equation App: Find the Width of a Frame (Factoring)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/AlIoxXQ-V50?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful\u2014when both sides of an equation are raised to an <i>even<\/i> power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.<\/p>\n<p>Completing the square is used to change a binomial of the form [latex]x^{2}+bx[\/latex] into a perfect square trinomial [latex]{{x}^{2}}+bx+{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex], which can be factored to [latex]{{\\left( x+\\frac{b}{2} \\right)}^{2}}[\/latex]. When solving quadratic equations by completing the square, be careful to add [latex]{{\\left( \\frac{b}{2} \\right)}^{2}}[\/latex] to <i>both<\/i> sides of the equation to maintain equality. The Square Root Property can then be used to solve for <i>x<\/i>. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.<\/p>\n<p>Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations, or any other quadratic equation! The Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex][\/latex].<\/p>\n<h2><\/h2>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1118\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Screenshot: Quadratic Equation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve Radical Equations - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qkZHKK77grM\">https:\/\/youtu.be\/qkZHKK77grM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Solve a Basic Radical Equation - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tT0Zwsto6AQ\">https:\/\/youtu.be\/tT0Zwsto6AQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 4: Solve Radical Equations - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/y2yHPfuL0Hs\">https:\/\/youtu.be\/y2yHPfuL0Hs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Completing the Square - Real Irrational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/IjCjbtrPWHM\">https:\/\/youtu.be\/IjCjbtrPWHM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Creating a Perfect Square Quadratic Trinomial Expression. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/vt-pM1LEP1M\">https:\/\/youtu.be\/vt-pM1LEP1M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solving Quadratic Equations Using Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4H5qZ_-8YM4\">https:\/\/youtu.be\/4H5qZ_-8YM4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Solving Quadratic Equations Using Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Fj-BP7uaWrI\">https:\/\/youtu.be\/Fj-BP7uaWrI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Quadratic Formula - One Real Rational Repeated Solution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/OXwwzWcxFgE\">https:\/\/youtu.be\/OXwwzWcxFgE<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Quadratic Equation App: Find the Width of a Frame (Factoring). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/AlIoxXQ-V50\">https:\/\/youtu.be\/AlIoxXQ-V50<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex2: Quadratic Formula - Two Real Irrational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tF0muV86dr0\">https:\/\/youtu.be\/tF0muV86dr0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Quadratic Formula - Two Real Rational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xtwO-n8lRPw\">https:\/\/youtu.be\/xtwO-n8lRPw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":20,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve Radical Equations - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/qkZHKK77grM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solve a Basic Radical Equation - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tT0Zwsto6AQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 4: Solve Radical Equations - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/y2yHPfuL0Hs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Completing the Square - Real Irrational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/IjCjbtrPWHM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Creating a Perfect Square Quadratic Trinomial Expression\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/vt-pM1LEP1M\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solving Quadratic Equations Using Square Roots\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/4H5qZ_-8YM4\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solving Quadratic Equations Using Square Roots\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Fj-BP7uaWrI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Quadratic Equation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Quadratic Formula - One Real Rational Repeated Solution\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/OXwwzWcxFgE\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Quadratic Equation App: Find the Width of a Frame (Factoring)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/AlIoxXQ-V50\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex2: Quadratic Formula - Two Real Irrational Solutions\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tF0muV86dr0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Quadratic Formula - Two Real Rational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xtwO-n8lRPw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"60313c10-a957-4321-994b-c119978babe4","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1118","chapter","type-chapter","status-publish","hentry"],"part":1053,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapters\/1118","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/wp\/v2\/users\/20"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapters\/1118\/revisions"}],"predecessor-version":[{"id":5988,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapters\/1118\/revisions\/5988"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/parts\/1053"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapters\/1118\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/wp\/v2\/media?parent=1118"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1118"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/wp\/v2\/contributor?post=1118"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/wp-json\/wp\/v2\/license?post=1118"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}