{"id":203,"date":"2023-11-08T16:10:24","date_gmt":"2023-11-08T16:10:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-or-watch-square-roots-and-completing-the-square\/"},"modified":"2026-02-05T11:37:17","modified_gmt":"2026-02-05T11:37:17","slug":"6-1-solving-quadratic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-1-solving-quadratic-equations\/","title":{"raw":"6.1 Solving Quadratic Equations","rendered":"6.1 Solving Quadratic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve quadratic equations using the square root property for equations with integer, rational, irrational, or complex number solutions.<\/li>\r\n \t<li>Solve quadratic equations by the method of completing the square for equations with integer, rational, irrational, or complex number solutions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nQuadratic equations can be solved using many methods. You should already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this section, you will learn two other ways to solve quadratic equations.\r\n<h2>Solve a Quadratic Equation by the Square Root Property<\/h2>\r\n<div style=\"font-weight: 400;\">\r\n\r\nIn this lesson we will solve quadratic equations where the variable is raised to the second power, [latex]x^2[\/latex] . Examples of the types of quadratic equations that will be solved include:\r\n<ul>\r\n \t<li>[latex]x^{2}=25[\/latex]<\/li>\r\n \t<li>[latex](x-3)^{2}=64[\/latex]<\/li>\r\n \t<li>[latex]3(x+1)^{2}=48[\/latex]<\/li>\r\n<\/ul>\r\nSome operations in mathematics have an inverse operation. The inverse operation of subtraction is addition. The inverse operation of multiplication is division. The inverse operation of taking the square root is squaring (raising to the power of 2). When we take the square root, we must remember to take both the positive AND the negative square roots.\r\n\r\n<\/div>\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 right side of the equation equal to 0.\r\n\r\n[latex]x^{2}-9=0[\/latex]\r\n\r\nNotice that the expression on the left can be factored; it is a difference of squares. Now you can solve this quadratic equation is by factoring the left side of the equation and then using the zero-product property.\r\n\r\n[latex]\\left(x+3\\right)\\left(x\u20133\\right)=0[\/latex].\r\n\r\nUsing the zero-product property, [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}=p[\/latex],\r\n\r\nthen [latex]x=\\sqrt{p}[\/latex] or [latex] x=-\\sqrt{p}[\/latex].\r\n\r\nThe property above states 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 [latex]p[\/latex] and the negative square root of [latex]p[\/latex].\r\n\r\n<\/div>\r\nWhen using the Square Root Property to solve an equation such as the one above ([latex]x^2=p[\/latex]), the solutions can be abbreviated as [latex]x=\\pm \\sqrt{p}[\/latex]. This is read as \"x equals plus or minus the square root of p.\" These still represent the same solutions of\u00a0[latex]x=\\sqrt{p}[\/latex] or [latex] x=-\\sqrt{p}[\/latex].\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 of the equation already has [latex]x^{2}[\/latex] isolated, you can take the square root of both sides to solve for [latex]x[\/latex].\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\nCheck your solutions:\r\n\r\n<strong>For [latex]x=3[\/latex]:<\/strong>\r\n\r\n[latex]\\begin{align}\\require{color}(\\color{Green}{3}\\color{black}{)^2-9} &amp;=0\\\\ 9-9 &amp;= 0\\\\ 0 &amp;=0 &amp;&amp; \\color{Green}{TRUE}\\end{align}[\/latex]\r\n\r\n<strong>For [latex]x=-3[\/latex]:<\/strong>\r\n\r\n[latex]\\begin{align}\\require{color}(\\color{Green}{-3}\\color{black}{)^2-9} &amp;=0\\\\ 9-9 &amp;= 0\\\\ 0 &amp;=0 &amp;&amp; \\color{Green}{TRUE}\\end{align}[\/latex]\r\n\r\nThey both solutions check. The solutions are\u00a0[latex]x=3[\/latex] or [latex]x=-3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that there is a difference here in solving [latex]x^{2}=9[\/latex]\u00a0and simplifying [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.\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\nWe need to isolate [latex]x^2[\/latex]. Let's do this by first subtracting [latex]5[\/latex] from both sides.\r\n<p style=\"text-align: center;\">[latex]10x^{2}+5=85[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\require{color}\\hspace{8mm}\\color{Green}{\\underline{-5}\\hspace{5mm}\\underline{-5}}[\/latex]<\/p>\r\nNext, let's divide by\u00a0[latex]10[\/latex] to isolate [latex]x^2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\dfrac{10x^{2}}{\\color{Green}{10}}=\\dfrac{80}{\\color{Green}{10}}[\/latex]<\/p>\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{align}{x}^{2} &amp;= 8\\\\ x &amp;= \\pm \\sqrt{8}\\\\ x &amp;= \\pm \\sqrt{(4)(2)}\\\\ x &amp;= \\pm \\sqrt{4}\\sqrt{2}\\\\ x &amp;= \\pm 2\\sqrt{2}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we have a quantity squared and that is what we need to isolate to solve this using the Square Root Property.\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\nStart by adding 50 to both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(x\u20132\\right)^{2}\u201350 &amp;= 0\\\\ \\left(x-2\\right)^{2} &amp;= 50\\end{align}[\/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{align}{\\left(x-2\\right)}^{2} &amp;= 50\\\\ \\sqrt{{\\left(x-2\\right)}^{2}} &amp;= \\sqrt{50}\\\\ x-2 &amp;= \\pm\\sqrt{50}\\end{align}[\/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.\r\n<p style=\"text-align: center;\">[latex] \\begin{align}x &amp;= 2\\pm \\sqrt{50}\\\\ x &amp;= 2\\pm \\sqrt{(25)(2)}\\\\ x &amp;= 2\\pm \\sqrt{25}\\sqrt{2}\\\\ x &amp;= 2\\pm 5\\sqrt{2}\\end{align}[\/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\nWhen you square a number, the product is called a perfect square. For example, [latex]5^2=5\\cdot 5=25[\/latex] so 25 is a perfect square. Here is a list of some integers that are also perfect squares: [latex]1,4,9,16,25,36,49,64,81,100...[\/latex]. We can also visualize integers that are perfect squares by arranging equal rows and columns to form a square, as shown below.\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-1-solving-quadratic-equations\/screenshot-2024-08-02-at-9-20-29%e2%80%afam\/\" rel=\"attachment wp-att-1704\"><img class=\"alignnone wp-image-1704 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-02-at-9.20.29\u202fAM.png\" alt=\"Grid representations of perfect squares: 9 equals 3 squared and 16 equals 4 squared.\" width=\"290\" height=\"241\" \/><\/a>\r\n\r\nTake a closer look at the perfect square trinomials below. What do you notice about the constant terms? How are those constant terms related to the factored form and the binomial squared?\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 33.333333%; text-align: center;\">\r\n<h3>Perfect Square Trinomial<\/h3>\r\n<\/td>\r\n<td style=\"width: 33.333333%; text-align: center;\">\r\n<h3>Factored Form<\/h3>\r\n<\/td>\r\n<td style=\"width: 33.333333%; text-align: center;\">\r\n<h3>Binomial Squared<\/h3>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2+10x+\\color{BurntOrange}{\\fbox{25}}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x+\\color{BurntOrange}{5}\\color{black}{)(x+}\\color{BurntOrange}{5}\\color{black}{)}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x+\\color{BurntOrange}{5}\\color{black}{)^2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2-6x+\\color{BurntOrange}{\\fbox{9}}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{3}\\color{black}{)(x-}\\color{BurntOrange}{3}\\color{black}{)}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{3}\\color{black}{)^2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2-8x+\\color{BurntOrange}{\\fbox{16}}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{4}\\color{black}{)(x-}\\color{BurntOrange}{4}\\color{black}{)}}[\/latex]<\/td>\r\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{4}\\color{black}{)^2}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDid you notice in the first example that the constant term, [latex]25[\/latex], of the trinomial is a perfect square? We know that [latex]25=5\\cdot 5=5^2[\/latex] and this helps us to confirm why there are [latex]5[\/latex]'s in the factored form and the [latex]5[\/latex] in the binomial squared form. Did you notice that the constant term of the trinomial is always positive, even when the [latex]x[\/latex] term is negative? Notice that this same idea applies to the other two examples above.\r\n\r\nLet's take a look at the [latex]x[\/latex] term (middle term) of the trinomial. How is it related to the binomial that is being squared? Looking at the first example above, did you notice that if you find half of the coefficient of the middle term of the trinomial, [latex]10[\/latex], that it is equal to [latex]5[\/latex]? Which is also the second term of the binomial squared. Does this same pattern hold true for the other above examples? Yes it does. In the second example, if we take half of [latex]-6[\/latex], we get [latex]-3[\/latex] and in the third example if we take half of [latex]-8[\/latex] we get [latex]-4[\/latex].\r\n\r\nNot all trinomials can be factored and not all trinomials in the form of [latex]ax^2+bx+c[\/latex] are perfect square trinomials. When they are not, we will need to change these trinomials into perfect square trinomials so that we can use this method to solve quadratic equations. 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. In this course, we will only be completing the square on trinomials that have a leading coefficient of [latex]1[\/latex]. In later courses you will learn how to complete the square when the leading coefficient does not equal 1.\r\n\r\nFirst, let's make sure we remember how to factor a perfect square trinomial.\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. Also notice\u00a0that the middle term (ignoring the sign) is twice the product of the square roots of these squared terms.\r\n\r\nBelow we have rewritten each term to make it easier to factor.\r\n\r\n[latex]9x^{2}=\\left(3x\\right)^{2}[\/latex]\r\n\r\n[latex]24x=2\\left(3x\\right)\\left(4\\right)[\/latex]\r\n\r\n[latex]16=4^{2}[\/latex]\r\n\r\nA trinomial in the form [latex]m^{2}-2mn+n^{2}[\/latex]\u00a0can be factored as\u00a0[latex](m-n)^{2}[\/latex].\r\n\r\nIn this case, the middle term is subtracted, so subtract [latex]m[\/latex] and [latex]n[\/latex] and square it to get\u00a0[latex](m\u2013n)^{2}[\/latex].\r\n\r\n[latex]\\begin{array}{c}\\,\\,\\,m=3x\\\\n=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\nIn this course when we complete the square, the leading coefficient, <em>a<\/em>, will equal 1. Let's look at an equation that does not have a perfect square trinomial.\r\n\r\n[latex]x^2+8x-84=0[\/latex]\r\n\r\nTo change the left side of the equation into a perfect square trinomial, let's start by adding 84 to both sides of the equation which is:\r\n\r\n[latex]x^2+8x=84[\/latex] and rewrite the equation to look like [latex]x^2+8x+\\color{BurntOrange}{\\_\\_\\_}\\color{black}{= 84+}\\color{BurntOrange}{\\_\\_\\_}[\/latex].\r\n\r\nNow we need to figure out what to put in the blanks to make the left side of the equation a perfect square trinomial. To do this, we will find half of the coefficient of the middle term which is [latex]\\dfrac{8}{2}=4[\/latex]. Then take that answer and square it [latex](4)^2=16[\/latex]. So, [latex]16[\/latex] is the number that we need to fill in the blanks with to make the left side into a perfect square trinomial.\r\n\r\n[latex]\\begin{align}x^2+8x+\\color{BurntOrange}{16} &amp;= 84+\\color{BurntOrange}{16}\\\\ x^2+8x+16 &amp;= 100\\end{align}[\/latex]\r\n\r\nNow, rewrite the left side of the equation as a binomial squared. Then square root both sides of the equation (using the Square Root Property) to solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{align}(x+4)^2 &amp;= 100\\\\ \\sqrt{(x+4)^2} &amp;= \\sqrt{100}\\\\ x+4 &amp;= \\pm 10\\end{align}[\/latex]\r\n\r\nNow write as two equations and solve each equation by subtracting [latex]4[\/latex] from both sides of each of the equations.\r\n\r\n[latex]x+4=10\\hspace{5mm}[\/latex] OR [latex]\\hspace{1cm}x+4=-10[\/latex]\r\n\r\n[latex]x=6\\hspace{1.5cm}[\/latex] OR [latex]\\hspace{1cm}x=-14[\/latex]\r\n\r\nYou might be wondering why this method is called Completing the Square. The following video will help you visualize geometrically what is meant by Completing the Square.\r\n\r\nhttps:\/\/youtu.be\/_YgUOsca7OU?si=xO4GcMBZBF8PMK1L&amp;t=14\r\n\r\nWe can use the following steps 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 [latex]b[\/latex] 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. In this case \u00a0we will add 4, then 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 trinomial.\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}}= \\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 exact solutions are\u00a0[latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex]. Sometimes you might be asked to give approximated solutions to a given place value. The approximated solutions rounded to the nearest hundredth are [latex]x\\approx 3.73, x\\approx 0.27[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\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\nWhat is [latex]b[\/latex]? [latex]\\hspace{5mm}b=-12[\/latex]\r\n\r\nNow take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] 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 exact solutions are [latex] x=6\\pm 2\\sqrt{10}[\/latex].\r\n\r\nOn this example, we have two irrational solutions.\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\nWhat is [latex]b[\/latex]? \u00a0 \u00a0[latex]b=-3[\/latex]\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] 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\nWe end up with two irrational solutions.\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 real solution.\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\r\nIn the next example, take notice of the type and number of solutions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve by completing the square. [latex]x^{2}-2x-41=-60[\/latex].\r\n\r\n[reveal-answer q=\"787113\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"787113\"]\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}-2x=-19[\/latex]<\/div>\r\nWhat is [latex]b[\/latex]? \u00a0 \u00a0[latex]b=-2[\/latex]\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-2\\right)=\\frac{-2}{2}=-1\\hfill \\\\ (-1)^{2}=1\\hfill \\end{array}[\/latex]<\/p>\r\nAdd the result to both sides of the equal sign.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x+1=-19+1\\\\x^{2}-2x+1=-18\\end{array}[\/latex]<\/p>\r\nRewrite the left side as a squared binomial. Then use the Square Root Property to solve.\r\n<p style=\"text-align: center;\">[latex]\\left(x-1\\right)^{2}=-18[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sqrt{(x-1)^{2}}=\\sqrt{-18}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] x-1=\\pm\\sqrt{-18}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] x-1=\\pm 3i \\sqrt{2}[\/latex]<\/p>\r\nSolve for [latex]x[\/latex] by adding\u00a0[latex]1[\/latex] to both sides. Simplify as needed.\r\n<p style=\"text-align: center;\">[latex]x=1\\pm 3i\\sqrt{2}[\/latex]<\/p>\r\nThe solutions can also be written as [latex]x=1\\pm 3\\sqrt{2}i[\/latex]. If you write it this way, make sure the [latex]i[\/latex] is not under the square root.\r\nThe exact solutions are [latex] x=1+ 3i\\sqrt{2}, x=1-3i\\sqrt{2}[\/latex].\r\n\r\nWe end up with two complex solutions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\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 we have a quadratic equation that cannot be factored, we can use completing the square to solve it. 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>Solve quadratic equations using the square root property for equations with integer, rational, irrational, or complex number solutions.<\/li>\n<li>Solve quadratic equations by the method of completing the square for equations with integer, rational, irrational, or complex number solutions.<\/li>\n<\/ul>\n<\/div>\n<p>Quadratic equations can be solved using many methods. You should already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this section, you will learn two other ways to solve quadratic equations.<\/p>\n<h2>Solve a Quadratic Equation by the Square Root Property<\/h2>\n<div style=\"font-weight: 400;\">\n<p>In this lesson we will solve quadratic equations where the variable is raised to the second power, [latex]x^2[\/latex] . Examples of the types of quadratic equations that will be solved include:<\/p>\n<ul>\n<li>[latex]x^{2}=25[\/latex]<\/li>\n<li>[latex](x-3)^{2}=64[\/latex]<\/li>\n<li>[latex]3(x+1)^{2}=48[\/latex]<\/li>\n<\/ul>\n<p>Some operations in mathematics have an inverse operation. The inverse operation of subtraction is addition. The inverse operation of multiplication is division. The inverse operation of taking the square root is squaring (raising to the power of 2). When we take the square root, we must remember to take both the positive AND the negative square roots.<\/p>\n<\/div>\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 right side of the equation equal to 0.<\/p>\n<p>[latex]x^{2}-9=0[\/latex]<\/p>\n<p>Notice that the expression on the left can be factored; it is a difference of squares. Now you can solve this quadratic equation is by factoring the left side of the equation and then using the zero-product property.<\/p>\n<p>[latex]\\left(x+3\\right)\\left(x\u20133\\right)=0[\/latex].<\/p>\n<p>Using the zero-product property, [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}=p[\/latex],<\/p>\n<p>then [latex]x=\\sqrt{p}[\/latex] or [latex]x=-\\sqrt{p}[\/latex].<\/p>\n<p>The property above states 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 [latex]p[\/latex] and the negative square root of [latex]p[\/latex].<\/p>\n<\/div>\n<p>When using the Square Root Property to solve an equation such as the one above ([latex]x^2=p[\/latex]), the solutions can be abbreviated as [latex]x=\\pm \\sqrt{p}[\/latex]. This is read as &#8220;x equals plus or minus the square root of p.&#8221; These still represent the same solutions of\u00a0[latex]x=\\sqrt{p}[\/latex] or [latex]x=-\\sqrt{p}[\/latex].<\/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 of the equation already has [latex]x^{2}[\/latex] isolated, you can take the square root of both sides to solve for [latex]x[\/latex].<\/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<p>Check your solutions:<\/p>\n<p><strong>For [latex]x=3[\/latex]:<\/strong><\/p>\n<p>[latex]\\begin{align}\\require{color}(\\color{Green}{3}\\color{black}{)^2-9} &=0\\\\ 9-9 &= 0\\\\ 0 &=0 && \\color{Green}{TRUE}\\end{align}[\/latex]<\/p>\n<p><strong>For [latex]x=-3[\/latex]:<\/strong><\/p>\n<p>[latex]\\begin{align}\\require{color}(\\color{Green}{-3}\\color{black}{)^2-9} &=0\\\\ 9-9 &= 0\\\\ 0 &=0 && \\color{Green}{TRUE}\\end{align}[\/latex]<\/p>\n<p>They both solutions check. The solutions are\u00a0[latex]x=3[\/latex] or [latex]x=-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 simplifying [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.<\/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>We need to isolate [latex]x^2[\/latex]. Let&#8217;s do this by first subtracting [latex]5[\/latex] from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]10x^{2}+5=85[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}\\hspace{8mm}\\color{Green}{\\underline{-5}\\hspace{5mm}\\underline{-5}}[\/latex]<\/p>\n<p>Next, let&#8217;s divide by\u00a0[latex]10[\/latex] to isolate [latex]x^2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{10x^{2}}{\\color{Green}{10}}=\\dfrac{80}{\\color{Green}{10}}[\/latex]<\/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{align}{x}^{2} &= 8\\\\ x &= \\pm \\sqrt{8}\\\\ x &= \\pm \\sqrt{(4)(2)}\\\\ x &= \\pm \\sqrt{4}\\sqrt{2}\\\\ x &= \\pm 2\\sqrt{2}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we have a quantity squared and that is what we need to isolate to solve this using the Square Root Property.<\/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>Start by adding 50 to both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(x\u20132\\right)^{2}\u201350 &= 0\\\\ \\left(x-2\\right)^{2} &= 50\\end{align}[\/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{align}{\\left(x-2\\right)}^{2} &= 50\\\\ \\sqrt{{\\left(x-2\\right)}^{2}} &= \\sqrt{50}\\\\ x-2 &= \\pm\\sqrt{50}\\end{align}[\/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.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x &= 2\\pm \\sqrt{50}\\\\ x &= 2\\pm \\sqrt{(25)(2)}\\\\ x &= 2\\pm \\sqrt{25}\\sqrt{2}\\\\ x &= 2\\pm 5\\sqrt{2}\\end{align}[\/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>When you square a number, the product is called a perfect square. For example, [latex]5^2=5\\cdot 5=25[\/latex] so 25 is a perfect square. Here is a list of some integers that are also perfect squares: [latex]1,4,9,16,25,36,49,64,81,100...[\/latex]. We can also visualize integers that are perfect squares by arranging equal rows and columns to form a square, as shown below.<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-1-solving-quadratic-equations\/screenshot-2024-08-02-at-9-20-29%e2%80%afam\/\" rel=\"attachment wp-att-1704\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1704 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-02-at-9.20.29\u202fAM.png\" alt=\"Grid representations of perfect squares: 9 equals 3 squared and 16 equals 4 squared.\" width=\"290\" height=\"241\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-02-at-9.20.29\u202fAM.png 290w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-02-at-9.20.29\u202fAM-65x54.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/Screenshot-2024-08-02-at-9.20.29\u202fAM-225x187.png 225w\" sizes=\"auto, (max-width: 290px) 100vw, 290px\" \/><\/a><\/p>\n<p>Take a closer look at the perfect square trinomials below. What do you notice about the constant terms? How are those constant terms related to the factored form and the binomial squared?<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 33.333333%; text-align: center;\">\n<h3>Perfect Square Trinomial<\/h3>\n<\/td>\n<td style=\"width: 33.333333%; text-align: center;\">\n<h3>Factored Form<\/h3>\n<\/td>\n<td style=\"width: 33.333333%; text-align: center;\">\n<h3>Binomial Squared<\/h3>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2+10x+\\color{BurntOrange}{\\fbox{25}}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x+\\color{BurntOrange}{5}\\color{black}{)(x+}\\color{BurntOrange}{5}\\color{black}{)}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x+\\color{BurntOrange}{5}\\color{black}{)^2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2-6x+\\color{BurntOrange}{\\fbox{9}}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{3}\\color{black}{)(x-}\\color{BurntOrange}{3}\\color{black}{)}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{3}\\color{black}{)^2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.333333%;\">[latex]\\large{x^2-8x+\\color{BurntOrange}{\\fbox{16}}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{4}\\color{black}{)(x-}\\color{BurntOrange}{4}\\color{black}{)}}[\/latex]<\/td>\n<td style=\"width: 33.333333%;\">[latex]\\large{(x-\\color{BurntOrange}{4}\\color{black}{)^2}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Did you notice in the first example that the constant term, [latex]25[\/latex], of the trinomial is a perfect square? We know that [latex]25=5\\cdot 5=5^2[\/latex] and this helps us to confirm why there are [latex]5[\/latex]&#8216;s in the factored form and the [latex]5[\/latex] in the binomial squared form. Did you notice that the constant term of the trinomial is always positive, even when the [latex]x[\/latex] term is negative? Notice that this same idea applies to the other two examples above.<\/p>\n<p>Let&#8217;s take a look at the [latex]x[\/latex] term (middle term) of the trinomial. How is it related to the binomial that is being squared? Looking at the first example above, did you notice that if you find half of the coefficient of the middle term of the trinomial, [latex]10[\/latex], that it is equal to [latex]5[\/latex]? Which is also the second term of the binomial squared. Does this same pattern hold true for the other above examples? Yes it does. In the second example, if we take half of [latex]-6[\/latex], we get [latex]-3[\/latex] and in the third example if we take half of [latex]-8[\/latex] we get [latex]-4[\/latex].<\/p>\n<p>Not all trinomials can be factored and not all trinomials in the form of [latex]ax^2+bx+c[\/latex] are perfect square trinomials. When they are not, we will need to change these trinomials into perfect square trinomials so that we can use this method to solve quadratic equations. 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. In this course, we will only be completing the square on trinomials that have a leading coefficient of [latex]1[\/latex]. In later courses you will learn how to complete the square when the leading coefficient does not equal 1.<\/p>\n<p>First, let&#8217;s make sure we remember how to factor a perfect square trinomial.<\/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. Also notice\u00a0that the middle term (ignoring the sign) is twice the product of the square roots of these squared terms.<\/p>\n<p>Below we have rewritten each term to make it easier to factor.<\/p>\n<p>[latex]9x^{2}=\\left(3x\\right)^{2}[\/latex]<\/p>\n<p>[latex]24x=2\\left(3x\\right)\\left(4\\right)[\/latex]<\/p>\n<p>[latex]16=4^{2}[\/latex]<\/p>\n<p>A trinomial in the form [latex]m^{2}-2mn+n^{2}[\/latex]\u00a0can be factored as\u00a0[latex](m-n)^{2}[\/latex].<\/p>\n<p>In this case, the middle term is subtracted, so subtract [latex]m[\/latex] and [latex]n[\/latex] and square it to get\u00a0[latex](m\u2013n)^{2}[\/latex].<\/p>\n<p>[latex]\\begin{array}{c}\\,\\,\\,m=3x\\\\n=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>In this course when we complete the square, the leading coefficient, <em>a<\/em>, will equal 1. Let&#8217;s look at an equation that does not have a perfect square trinomial.<\/p>\n<p>[latex]x^2+8x-84=0[\/latex]<\/p>\n<p>To change the left side of the equation into a perfect square trinomial, let&#8217;s start by adding 84 to both sides of the equation which is:<\/p>\n<p>[latex]x^2+8x=84[\/latex] and rewrite the equation to look like [latex]x^2+8x+\\color{BurntOrange}{\\_\\_\\_}\\color{black}{= 84+}\\color{BurntOrange}{\\_\\_\\_}[\/latex].<\/p>\n<p>Now we need to figure out what to put in the blanks to make the left side of the equation a perfect square trinomial. To do this, we will find half of the coefficient of the middle term which is [latex]\\dfrac{8}{2}=4[\/latex]. Then take that answer and square it [latex](4)^2=16[\/latex]. So, [latex]16[\/latex] is the number that we need to fill in the blanks with to make the left side into a perfect square trinomial.<\/p>\n<p>[latex]\\begin{align}x^2+8x+\\color{BurntOrange}{16} &= 84+\\color{BurntOrange}{16}\\\\ x^2+8x+16 &= 100\\end{align}[\/latex]<\/p>\n<p>Now, rewrite the left side of the equation as a binomial squared. Then square root both sides of the equation (using the Square Root Property) to solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{align}(x+4)^2 &= 100\\\\ \\sqrt{(x+4)^2} &= \\sqrt{100}\\\\ x+4 &= \\pm 10\\end{align}[\/latex]<\/p>\n<p>Now write as two equations and solve each equation by subtracting [latex]4[\/latex] from both sides of each of the equations.<\/p>\n<p>[latex]x+4=10\\hspace{5mm}[\/latex] OR [latex]\\hspace{1cm}x+4=-10[\/latex]<\/p>\n<p>[latex]x=6\\hspace{1.5cm}[\/latex] OR [latex]\\hspace{1cm}x=-14[\/latex]<\/p>\n<p>You might be wondering why this method is called Completing the Square. The following video will help you visualize geometrically what is meant by Completing the Square.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"How to really complete the square for solving a quadratic equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_YgUOsca7OU?start=14&#38;feature=oembed\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We can use the following steps 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 [latex]b[\/latex] 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. In this case \u00a0we will add 4, then 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 trinomial.\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}}= \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The exact solutions are\u00a0[latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex]. Sometimes you might be asked to give approximated solutions to a given place value. The approximated solutions rounded to the nearest hundredth are [latex]x\\approx 3.73, x\\approx 0.27[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\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>What is [latex]b[\/latex]? [latex]\\hspace{5mm}b=-12[\/latex]<\/p>\n<p>Now take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] 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 exact solutions are [latex]x=6\\pm 2\\sqrt{10}[\/latex].<\/p>\n<p>On this example, we have two irrational solutions.<\/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>What is [latex]b[\/latex]? \u00a0 \u00a0[latex]b=-3[\/latex]<br \/>\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] 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<p>We end up with two irrational solutions.<\/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 real solution.<\/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<p>In the next example, take notice of the type and number of solutions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve by completing the square. [latex]x^{2}-2x-41=-60[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787113\">Show Solution<\/span><\/p>\n<div id=\"q787113\" 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}-2x=-19[\/latex]<\/div>\n<p>What is [latex]b[\/latex]? \u00a0 \u00a0[latex]b=-2[\/latex]<br \/>\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-2\\right)=\\frac{-2}{2}=-1\\hfill \\\\ (-1)^{2}=1\\hfill \\end{array}[\/latex]<\/p>\n<p>Add the result to both sides of the equal sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x+1=-19+1\\\\x^{2}-2x+1=-18\\end{array}[\/latex]<\/p>\n<p>Rewrite the left side as a squared binomial. Then use the Square Root Property to solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x-1\\right)^{2}=-18[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{(x-1)^{2}}=\\sqrt{-18}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x-1=\\pm\\sqrt{-18}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x-1=\\pm 3i \\sqrt{2}[\/latex]<\/p>\n<p>Solve for [latex]x[\/latex] by adding\u00a0[latex]1[\/latex] to both sides. Simplify as needed.<\/p>\n<p style=\"text-align: center;\">[latex]x=1\\pm 3i\\sqrt{2}[\/latex]<\/p>\n<p>The solutions can also be written as [latex]x=1\\pm 3\\sqrt{2}i[\/latex]. If you write it this way, make sure the [latex]i[\/latex] is not under the square root.<br \/>\nThe exact solutions are [latex]x=1+ 3i\\sqrt{2}, x=1-3i\\sqrt{2}[\/latex].<\/p>\n<p>We end up with two complex solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/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 we have a quadratic equation that cannot be factored, we can use completing the square to solve it. 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-203\">\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":395986,"menu_order":1,"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 - 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