{"id":16250,"date":"2020-02-12T17:42:04","date_gmt":"2020-02-12T17:42:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/?post_type=chapter&#038;p=16250"},"modified":"2020-02-12T17:42:04","modified_gmt":"2020-02-12T17:42:04","slug":"square-roots-and-completing-the-square","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/chapter\/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<\/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\nThe property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of<i> a<\/i> and the negative square root of <i>a<\/i>.\r\n\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\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\nFirst, let us make sure we can recognize a perfect square trinomial and how to factor it.\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. If you do not start with a perfect square trinomial, you can complete the square to make what you have into one.\r\n\r\nTo 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\">\r\n<h3>Steps for Completing The Square<\/h3>\r\nWe will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step.\r\n<ol>\r\n \t<li>Given a quadratic equation that cannot be factored and with [latex]a=1[\/latex], first add or subtract the constant term to the right side of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/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]\\begin{array}{c}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The left side of the equation can now be factored as a perfect square.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Use the square root property and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve by completing the square.\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\"]\r\n\r\nSince you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.\r\n\r\nFirst, move the constant term to the right side of the equal sign.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-12x=4\\end{array}[\/latex]<\/p>\r\nIdentify [latex]b[\/latex]:\u00a0[latex]b=-12[\/latex]\r\n\r\nThen take [latex]\\frac{1}{2}[\/latex] of the <em>b <\/em>term and square it. 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 will miss one of the solutions.\r\n<p style=\"text-align: center;\">[latex] x-6=\\pm\\sqrt{40}[\/latex]<\/p>\r\nSolve for [latex]x[\/latex] by adding\u00a0[latex]6[\/latex] to both sides. Simplify as needed.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x &amp; =6\\pm \\sqrt{40}\\\\ &amp; =6\\pm \\sqrt{4}\\sqrt{10}\\\\ &amp; =6\\pm 2\\sqrt{10}\\end{array}[\/latex]<\/p>\r\nThe answer is [latex] x=6\\pm 2\\sqrt{10}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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 rational.\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<\/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>The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of<i> a<\/i> and the negative square root of <i>a<\/i>.<\/p>\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<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>First, let us make sure we can recognize a perfect square trinomial and how to factor it.<\/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. If you do not start with a perfect square trinomial, you can complete the square to make what you have into one.<\/p>\n<p>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\">\n<h3>Steps for Completing The Square<\/h3>\n<p>We will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step.<\/p>\n<ol>\n<li>Given a quadratic equation that cannot be factored and with [latex]a=1[\/latex], first add or subtract the constant term to the right side of the equal sign.\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/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]\\begin{array}{c}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have:\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The left side of the equation can now be factored as a perfect square.\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Use the square root property and solve.\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve by completing the square.\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\">\n<p>Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.<\/p>\n<p>First, move the constant term to the right side of the equal sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-12x=4\\end{array}[\/latex]<\/p>\n<p>Identify [latex]b[\/latex]:\u00a0[latex]b=-12[\/latex]<\/p>\n<p>Then take [latex]\\frac{1}{2}[\/latex] of the <em>b <\/em>term and square it. 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 will miss one of the solutions.<\/p>\n<p style=\"text-align: center;\">[latex]x-6=\\pm\\sqrt{40}[\/latex]<\/p>\n<p>Solve for [latex]x[\/latex] by adding\u00a0[latex]6[\/latex] 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<p>The answer is [latex]x=6\\pm 2\\sqrt{10}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\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-3\" 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 rational.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" 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","protected":false},"author":160,"menu_order":26,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16250","chapter","type-chapter","status-publish","hentry"],"part":16105,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapters\/16250","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/wp\/v2\/users\/160"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapters\/16250\/revisions"}],"predecessor-version":[{"id":16251,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapters\/16250\/revisions\/16251"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/parts\/16105"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapters\/16250\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/wp\/v2\/media?parent=16250"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=16250"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/wp\/v2\/contributor?post=16250"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-prealgebra\/wp-json\/wp\/v2\/license?post=16250"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}