{"id":16442,"date":"2019-10-03T15:48:48","date_gmt":"2019-10-03T15:48:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-square-roots-and-completing-the-square\/"},"modified":"2024-05-02T15:47:23","modified_gmt":"2024-05-02T15:47:23","slug":"read-or-watch-square-roots-and-completing-the-square","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-square-roots-and-completing-the-square\/","title":{"raw":"Square Roots and Completing the Square","rendered":"Square Roots and Completing the Square"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the square root\u00a0property to solve a quadratic equation<\/li>\r\n \t<li>Complete the square to solve a quadratic equation with irrational roots<\/li>\r\n<\/ul>\r\n<\/div>\r\nQuadratic equations can be solved using many methods. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this section, you will use square roots to learn another way to solve quadratic equations\u2014and this method will work with <i>all<\/i> quadratic equations.\r\n<h2>Solve a Quadratic Equation by the Square Root Property<\/h2>\r\nOne way to solve the quadratic equation [latex]x^{2}=9[\/latex]\u00a0is to subtract\u00a0[latex]9[\/latex] from both sides to get one side equal to 0: [latex]x^{2}-9=0[\/latex]. The expression on the left can be factored; it is a difference of squares:\u00a0[latex]\\left(x+3\\right)\\left(x\u20133\\right)=0[\/latex]. Using the zero factor property, you know this means [latex]x+3=0[\/latex] or [latex]x\u20133=0[\/latex], so [latex]x=\u22123[\/latex] or\u00a0[latex]3[\/latex].\r\n\r\nAnother property that would let you solve this equation more easily is called the square root property.\r\n<div class=\"textbox shaded\">\r\n<h3>The Square Root Property<\/h3>\r\nIf [latex]x^{2}=a[\/latex], then [latex] x=\\sqrt{a}[\/latex] or [latex] -\\sqrt{a}[\/latex].\r\n\r\n&nbsp;\r\n\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, [latex]a &gt; 0[\/latex], [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, [latex]a &lt; 0[\/latex], [latex]x^2=a[\/latex] has no solutions.<\/li>\r\n \t<li>When a is zero, [latex]a = 0[\/latex], [latex]x^2=a[\/latex] has one solution: [latex]a = 0[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nA shortcut way to write \u201c[latex] \\sqrt{a}[\/latex]\u201d or \u201c[latex] -\\sqrt{a}[\/latex]\u201d 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\"]\r\n\r\nSince 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. Do not forget to use both positive and negative square roots!\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}=9 \\\\ x=\\pm\\sqrt{9} \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\pm3[\/latex] (that is, [latex]x=3[\/latex] or [latex]-3[\/latex])<\/p>\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\u00a0[latex]9[\/latex]. 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\r\nIn our first video, we will show more examples of using the square root property to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/Fj-BP7uaWrI\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\"]\r\n\r\nIf 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 cannot simplify that. Subtract\u00a0[latex]5[\/latex] from both sides to get the [latex]x^{2}[\/latex]\u00a0term by itself.\r\n<p style=\"text-align: center;\">[latex]10x^{2}=80[\/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\u00a0[latex]10[\/latex] before you take the square root will be a little easier.\r\n<p style=\"text-align: center;\">[latex]x^{2}=8[\/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}{ll}{{x}^{2}} &amp; =8\\\\ x &amp; =\\pm \\sqrt{8}\\\\ &amp; =\\pm \\sqrt{(4)(2)}\\\\ &amp; =\\pm \\sqrt{4}\\sqrt{2}\\\\ &amp; =\\pm 2\\sqrt{2}\\end{array}[\/latex]<\/p>\r\nThe answer is [latex] x=\\pm 2\\sqrt{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]197330[\/ohm_question]\r\n\r\n<\/div>\r\nSometimes more than just the [latex]x[\/latex] 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\"]\r\n\r\nAgain, taking the square root of both sides at this stage will leave something you cannot 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[\/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 [latex]x[\/latex] on the left, you need to add\u00a0[latex]2[\/latex] to both sides.\r\n\r\nBe sure to simplify the radical if possible.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{ll}x &amp; =2\\pm \\sqrt{50} \\\\ &amp; =2\\pm \\sqrt{(25)(2)} \\\\ &amp; =2\\pm \\sqrt{25}\\sqrt{2} \\\\ &amp; =2\\pm 5\\sqrt{2}\\end{array}[\/latex]<\/p>\r\nThe answer is [latex] x=2\\pm 5\\sqrt{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, you will see more examples of using square roots to solve quadratic equations.\r\n\r\nhttps:\/\/youtu.be\/4H5qZ_-8YM4\r\n<h2><\/h2>\r\n<h2>Solve a Quadratic Equation by Completing the Square<\/h2>\r\nNot all quadratic equations can be factored or solved in their original form using the square root property. In these cases, we may use a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property.\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=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]9x^{2}\u201324x+16[\/latex].\r\n[reveal-answer q=\"290635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"290635\"]\r\n\r\nFirst notice that the [latex]x^{2}[\/latex]\u00a0term and the constant term are both perfect squares.\r\n\r\n[latex]\\begin{array}{l}9x^{2}=\\left(3x\\right)^{2} \\\\ 16=4^{2}\\end{array}[\/latex]\r\n\r\nThen notice that the middle term (ignoring the sign) is twice the product of the square roots of these squared terms.\r\n\r\n[latex]24x=2\\left(3x\\right)\\left(4\\right)[\/latex]\r\n\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\r\n[latex]\\begin{array}{c}\\,\\,\\,r=3x\\\\s=4\\\\9x^{2}-24x+16=\\left(3x-4\\right)^{2}\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf this were an equation, we could solve using either the square root property or the zero product property.\r\n\r\nYou can use the same 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 [latex]0[\/latex].\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. This results in 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 term 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 (and it will still work even when you can factor!).\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\nNotice that, to complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n<div class=\"textbox shaded\">\r\n<h3>Steps for Completing The Square<\/h3>\r\n<ol>\r\n \t<li>Given a quadratic equation with [latex]a=1[\/latex], first add or subtract the constant term c to the right side of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\r\n<div style=\"text-align: center;\">[latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex]<\/div><\/li>\r\n \t<li style=\"text-align: left;\">Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side<\/li>\r\n \t<li>The left side of the equation can now be factored as a perfect square.\u00a0\u00a0[latex]x^{2}+bx+\\left(\\frac{b}{2}\\right)^{2}=\\left(x+\\frac{b}{2}\\right)^{2}[\/latex]<\/li>\r\n \t<li>Use the square root property to solve.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]147603[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve by completing the square: [latex]{x}^{2}-3x - 5=0[\/latex].\r\n[reveal-answer q=\"567568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567568\"]\r\n\r\nFirst, move the constant term to the right side of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\r\nIdentify [latex]b[\/latex]:\u00a0 \u00a0[latex]b=-3[\/latex]\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the <em>b <\/em>term and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nAdd the result to both sides of the equal sign.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nFactor the left side as a perfect square and simplify the right side.\r\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\r\nUse the square root property and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}\\hfill &amp; = \\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ x-\\frac{3}{2} &amp; =\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x &amp; =\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=\\frac{3+\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3-\\sqrt{29}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, you will see more examples of how to use completing the square to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/PsbYUySRjFo\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 is slightly different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve by completing the square. [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\"]\r\n\r\nRewrite 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.\u00a0[latex]0[\/latex] has only one root.\r\n<p style=\"text-align: center;\">[latex]x+8=0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=-8[\/latex]<\/p>\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\u00a0[latex]47[\/latex] 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\u00a0[latex]16[\/latex]).\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\r\nIn our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are irrational.\r\n\r\nhttps:\/\/youtu.be\/IjCjbtrPWHM\r\n<h2>Summary<\/h2>\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 [latex]x[\/latex]. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the square root\u00a0property to solve a quadratic equation<\/li>\n<li>Complete the square to solve a quadratic equation with irrational roots<\/li>\n<\/ul>\n<\/div>\n<p>Quadratic equations can be solved using many methods. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this section, you will use square roots to learn another way to solve quadratic equations\u2014and this method will work with <i>all<\/i> quadratic equations.<\/p>\n<h2>Solve a Quadratic Equation by the Square Root Property<\/h2>\n<p>One way to solve the quadratic equation [latex]x^{2}=9[\/latex]\u00a0is to subtract\u00a0[latex]9[\/latex] from both sides to get one side equal to 0: [latex]x^{2}-9=0[\/latex]. The expression on the left can be factored; it is a difference of squares:\u00a0[latex]\\left(x+3\\right)\\left(x\u20133\\right)=0[\/latex]. Using the zero factor property, you know this means [latex]x+3=0[\/latex] or [latex]x\u20133=0[\/latex], so [latex]x=\u22123[\/latex] or\u00a0[latex]3[\/latex].<\/p>\n<p>Another property that would let you solve this equation more easily is called the square root property.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Square Root Property<\/h3>\n<p>If [latex]x^{2}=a[\/latex], then [latex]x=\\sqrt{a}[\/latex] or [latex]-\\sqrt{a}[\/latex].<\/p>\n<p>&nbsp;<\/p>\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, [latex]a > 0[\/latex], [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, [latex]a < 0[\/latex], [latex]x^2=a[\/latex] has no solutions.<\/li>\n<li>When a is zero, [latex]a = 0[\/latex], [latex]x^2=a[\/latex] has one solution: [latex]a = 0[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>A shortcut way to write \u201c[latex]\\sqrt{a}[\/latex]\u201d or \u201c[latex]-\\sqrt{a}[\/latex]\u201d 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\">\n<p>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. Do not forget to use both positive and negative square roots!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^{2}=9 \\\\ x=\\pm\\sqrt{9} \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[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\u00a0[latex]9[\/latex]. 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<p>In our first video, we will show more examples of using the square root property to solve a quadratic equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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<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\">\n<p>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 cannot simplify that. Subtract\u00a0[latex]5[\/latex] from both sides to get the [latex]x^{2}[\/latex]\u00a0term by itself.<\/p>\n<p style=\"text-align: center;\">[latex]10x^{2}=80[\/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\u00a0[latex]10[\/latex] before you take the square root will be a little easier.<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}=8[\/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}{ll}{{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<p>The answer is [latex]x=\\pm 2\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm197330\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=197330&theme=oea&iframe_resize_id=ohm197330&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Sometimes more than just the [latex]x[\/latex] 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\">\n<p>Again, taking the square root of both sides at this stage will leave something you cannot 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[\/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 [latex]x[\/latex] on the left, you need to add\u00a0[latex]2[\/latex] to both sides.<\/p>\n<p>Be sure to simplify the radical if possible.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}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<p>The answer is [latex]x=2\\pm 5\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, you will see more examples of using square roots to solve quadratic equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<h2><\/h2>\n<h2>Solve a Quadratic Equation by Completing the Square<\/h2>\n<p>Not all quadratic equations can be factored or solved in their original form using the square root property. In these cases, we may use a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property.<\/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=\"textbox exercises\">\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=\"q290635\">Show Solution<\/span><\/p>\n<div id=\"q290635\" class=\"hidden-answer\" style=\"display: none\">\n<p>First notice that the [latex]x^{2}[\/latex]\u00a0term and the constant term are both perfect squares.<\/p>\n<p>[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 these squared terms.<\/p>\n<p>[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>[latex]\\begin{array}{c}\\,\\,\\,r=3x\\\\s=4\\\\9x^{2}-24x+16=\\left(3x-4\\right)^{2}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>If this were an equation, we could solve using either the square root property or the zero product property.<\/p>\n<p>You can use the same 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 [latex]0[\/latex].<\/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. This results in 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 term that will make a trinomial a perfect square.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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 (and it will still work even when you can factor!).<\/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>Notice that, to complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\n<div class=\"textbox shaded\">\n<h3>Steps for Completing The Square<\/h3>\n<ol>\n<li>Given a quadratic equation with [latex]a=1[\/latex], first add or subtract the constant term c to the right side of the equal sign.\n<div style=\"text-align: center;\">[latex]{x}^{2}+bx=-c[\/latex]<\/div>\n<\/li>\n<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\n<div style=\"text-align: center;\">[latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex]<\/div>\n<\/li>\n<li style=\"text-align: left;\">Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side<\/li>\n<li>The left side of the equation can now be factored as a perfect square.\u00a0\u00a0[latex]x^{2}+bx+\\left(\\frac{b}{2}\\right)^{2}=\\left(x+\\frac{b}{2}\\right)^{2}[\/latex]<\/li>\n<li>Use the square root property to solve.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm147603\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=147603&theme=oea&iframe_resize_id=ohm147603&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve by completing the square: [latex]{x}^{2}-3x - 5=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567568\">Show Solution<\/span><\/p>\n<div id=\"q567568\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, move the constant term to the right side of the equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\n<p>Identify [latex]b[\/latex]:\u00a0 \u00a0[latex]b=-3[\/latex]<br \/>\nThen, take [latex]\\frac{1}{2}[\/latex] of the <em>b <\/em>term and square it.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Add the result to both sides of the equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Factor the left side as a perfect square and simplify the right side.<\/p>\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\n<p>Use the square root property and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}\\hfill & = \\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ x-\\frac{3}{2} & =\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x & =\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=\\frac{3+\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3-\\sqrt{29}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, you will see more examples of how to use completing the square to solve a quadratic equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 1:  Completing the Square - Real Rational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/PsbYUySRjFo?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 is slightly different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve by completing the square. [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\">\n<p>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.\u00a0[latex]0[\/latex] has only one root.<\/p>\n<p style=\"text-align: center;\">[latex]x+8=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[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\u00a0[latex]47[\/latex] 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\u00a0[latex]16[\/latex]).<\/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<p>In our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are irrational.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" 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<h2>Summary<\/h2>\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 [latex]x[\/latex]. 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\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-16442\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at: http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Ex 1: Solving Quadratic Equations Using Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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 2: Solving Quadratic Equations Using Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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: Completing the Square - Real Rational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/PsbYUySRjFo\">https:\/\/youtu.be\/PsbYUySRjFo<\/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) for Lumen Learning. <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><\/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":169554,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at: http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solving Quadratic Equations Using Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Fj-BP7uaWrI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solving Quadratic Equations Using Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/4H5qZ_-8YM4\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Completing the Square - Real Rational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/PsbYUySRjFo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Completing the Square - Real Irrational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/IjCjbtrPWHM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"4f34e13f8ee047d5b04458cd40046ecf, a91ce3a0d22b44e4a4e4e3847c533835, f9ae0b43b12946cd9f4222605af5f934","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16442","chapter","type-chapter","status-publish","hentry"],"part":16201,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16442","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":12,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16442\/revisions"}],"predecessor-version":[{"id":20508,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16442\/revisions\/20508"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/16201"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16442\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16442"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16442"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16442"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}