{"id":18335,"date":"2022-04-15T19:19:02","date_gmt":"2022-04-15T19:19:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=18335"},"modified":"2022-04-28T23:16:30","modified_gmt":"2022-04-28T23:16:30","slug":"cr-14-solving-quadratic-equations-using-factoring-square-root-method-completing-the-square-and-quadratic-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/cr-14-solving-quadratic-equations-using-factoring-square-root-method-completing-the-square-and-quadratic-formula\/","title":{"raw":"CR.14: Solving Quadratic Equations Using Factoring, Square Root Method, Completing the Square, and Quadratic Formula","rendered":"CR.14: Solving Quadratic Equations Using Factoring, Square Root Method, Completing the Square, and Quadratic Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a quadratic equation to solve it.<\/li>\r\n \t<li>Use the square root property to solve a quadratic equation.<\/li>\r\n \t<li>Complete the square to solve a quadratic equation.<\/li>\r\n \t<li>Use the quadratic formula to solve a quadratic equation.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: identifying polynomials<\/h3>\r\nRecall that a polynomial is an expression containing variable terms joined together by addition or subtraction. We can identify the\u00a0<strong>degree<\/strong> of the polynomial by finding the term having the highest power (largest exponent on its variable). That term will be the\u00a0<strong>leading term<\/strong> of the polynomial and its degree is the degree of the polynomial.\r\n\r\n<\/div>\r\nAn equation containing a second-degree polynomial is called a <strong>quadratic equation<\/strong>. For example, equations such as [latex]2{x}^{2}+3x - 1=0[\/latex] and [latex]{x}^{2}-4=0[\/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.\r\n\r\nOften the easiest method of solving a quadratic equation is <strong>factoring<\/strong>. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.\r\n\r\nIf a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if [latex]a\\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where <em>a <\/em>and <em>b <\/em>are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.\r\n\r\nMultiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression [latex]\\left(x - 2\\right)\\left(x+3\\right)[\/latex] by multiplying the two factors together.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)\\hfill&amp;={x}^{2}+3x - 2x - 6\\hfill \\\\ \\hfill&amp;={x}^{2}+x - 6\\hfill \\end{array}[\/latex]<\/div>\r\nThe product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[\/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.\r\n\r\nThe process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], where <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex]. The equation [latex]{x}^{2}+x - 6=0[\/latex] is in standard form.\r\n<div class=\"textbox examples\">\r\n<h3>recall the greatest common factor<\/h3>\r\nThe <strong>greatest common factor<\/strong> (GCF) is the largest number or variable expression that can divided evenly into two or more numbers or expressions.\r\n\r\nTo find the GCF of an expression containing variable terms, first find the GCF of the coefficients, then find the GCF of the variables. The GCF of the variables will be the smallest degree of each variable that appears in each term.\r\n\r\nFor example,\u00a0[latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest expression that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].\r\n\r\nWhen factoring, remember to look first for a greatest common factor. If you can find one to factor out, it will usually make your factoring task less challenging.\r\n\r\n<\/div>\r\nWe can use the zero-product property to solve quadratic equations in which we first have to factor out the <strong>greatest common factor<\/strong> (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Zero-Product Property and Quadratic Equations<\/h3>\r\nThe <strong>zero-product property<\/strong> states\r\n<div style=\"text-align: center;\">[latex]\\text{If }a\\cdot b=0,\\text{ then }a=0\\text{ or }b=0[\/latex],<\/div>\r\nwhere <em>a <\/em>and <em>b <\/em>are real numbers or algebraic expressions.\r\n\r\nA <strong>quadratic equation<\/strong> is an equation containing a second-degree polynomial; for example\r\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx+c=0[\/latex]<\/div>\r\nwhere <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers, and [latex]a\\ne 0[\/latex]. It is in standard form.\r\n\r\n<\/div>\r\n<h2>Solving Quadratics with a Leading Coefficient of 1<\/h2>\r\nIn the quadratic equation [latex]{x}^{2}+x - 6=0[\/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[\/latex], is 1. We have one method of factoring quadratic equations in this form.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic equation with the leading coefficient of 1, factor it<\/h3>\r\n<ol>\r\n \t<li>Find two numbers whose product equals <em>c<\/em> and whose sum equals <em>b<\/em>.<\/li>\r\n \t<li>Use those numbers to write two factors of the form [latex]\\left(x+k\\right)\\text{ or }\\left(x-k\\right)[\/latex], where <em>k <\/em>is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and [latex]-2[\/latex], the factors are [latex]\\left(x+1\\right)\\left(x - 2\\right)[\/latex].<\/li>\r\n \t<li>Solve using the zero-product property by setting each factor equal to zero and solving for the variable.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring and Solving a Quadratic with Leading Coefficient of 1<\/h3>\r\nFactor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].\r\n\r\n[reveal-answer q=\"16111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16111\"]\r\n\r\nTo factor [latex]{x}^{2}+x - 6=0[\/latex], we look for two numbers whose product equals [latex]-6[\/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}1\\cdot \\left(-6\\right)\\hfill \\\\ \\left(-6\\right)\\cdot 1\\hfill \\\\ 2\\cdot \\left(-3\\right)\\hfill \\\\ 3\\cdot \\left(-2\\right)\\hfill \\end{array}[\/latex]<\/div>\r\nThe last pair, [latex]3\\cdot \\left(-2\\right)[\/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.\r\n<div style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/div>\r\nTo solve this equation, we use the zero-product property. Set each factor equal to zero and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/div>\r\nThe two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/latex]. We can see how the solutions relate to the graph below. The solutions are the <em>x-<\/em>intercepts of the graph of [latex]{x}^{2}+x - 6[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24224527\/CNX_CAT_Figure_02_05_002.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well.\" width=\"487\" height=\"588\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor and solve the quadratic equation: [latex]{x}^{2}-5x - 6=0[\/latex].\r\n\r\n[reveal-answer q=\"220537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"220537\"]\r\n\r\n[latex]\\left(x - 6\\right)\\left(x+1\\right)=0;x=6,x=-1[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]2029[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using Other Factoring Methods<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Setting it Equal to Zero and Factoring<\/h3>\r\nFactor and solve the equation: [latex]{x}^{3}=100{x}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"16120\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16120\"]\r\n\r\nFirst we must get everything equal to zero. If we divide both sides by x instead, we may miss one of our answers. That is why we must first set the equation equal to zero:\r\n[latex]{x}^{3}-100{x}^{2}=0[\/latex]\r\n\r\nNext we will factor out the GCF: [latex]{x}^{2}(x-100)=0[\/latex].\r\n\r\nTo solve this equation, we use the zero-product property. Set each factor equal to zero and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {x}^{2}(x-100)=0\\hfill \\\\x^2=0 \\hfill\\\\ x=0\\hfill \\\\ (x - 100)=0\\hfill \\\\ x=100\\hfill \\end{array}[\/latex]<\/div>\r\nThe two solutions are [latex]x=0[\/latex] and [latex]x=100[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor and solve the quadratic equation: [latex]{x}^{20}=2{x}^{19}[\/latex].\r\n\r\n[reveal-answer q=\"220540\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"220540\"]\r\n\r\n[latex]{x}^{19}\\left(x - 2\\right)=0;x=0,x=2[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]248381[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Method of Factor by Grouping<\/h3>\r\nFactor and solve the equation: [latex]{x}^{3}+11{x}^{2}-121x-1331=0[\/latex].\r\n\r\n[reveal-answer q=\"16130\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16130\"]\r\n\r\nWe will solve this we will use the the method of factoring by grouping. First group the terms.\r\n[latex]{x}^{3}+11{x}^{2}-121x-1331=0[\/latex]\r\n[latex]({x}^{3}+11{x}^{2})-(121x+1331)=0[\/latex]\r\n\r\nThe greatest common factor in the first grouping is [latex]{x}^{2}[\/latex].\r\nThe greatest common factor in the second grouping is [latex]121[\/latex].\r\n\r\nFactor out [latex]{x}^{2}[\/latex] from the first grouping and [latex]121[\/latex] from the second grouping.\r\n[latex]({x}^{3}+11{x}^{2})-(121x+1331)=0[\/latex]\r\n[latex]{x}^{2}(x+11)-121(x+11)=0[\/latex]\r\n\r\nThe common factor in the [latex]{x}^{2}(x+11)-121(x+11)=0[\/latex] is [latex](x+11)[\/latex]. We will now factor this out:\r\n[latex](x+11)({x}^{2}-121)=0[\/latex]\r\n\r\nWe can factor one more time:\r\n[latex](x+11)(x+11)(x-11)=0[\/latex]\r\n\r\nTo solve this equation, we use the zero-product property. Set each factor equal to zero and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} (x+11)(x+11)(x-11)=0\\hfill \\\\(x+11)=0 \\hfill\\\\ x=-11\\hfill \\\\ (x +11)=0\\hfill \\\\ x=-11\\hfill \\\\(x-11)=0 \\hfill\\\\ x=11\\hfill \\end{array}[\/latex]<\/div>\r\nOne solution is repeated twice, but we only need to write it once.\r\nTherefore, the two solutions are [latex]x=-11[\/latex] and [latex]x=11[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor and solve the quadratic equation: [latex]{x}^{3}-12{x}^2-144x+1728=0[\/latex].\r\n\r\n[reveal-answer q=\"220550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"220550\"]\r\n\r\n[latex]{x}^{3}-12{x}^2-144x+1728=0;x=-12,x=12[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]248391[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using the Square Root Property<\/h2>\r\nWhen there is no linear term in the equation, another method of solving a quadratic equation is by using the <strong>square root property<\/strong>, in which we isolate the [latex]{x}^{2}[\/latex] term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the [latex]{x}^{2}[\/latex] term so that the square root property can be used.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Square Root Property<\/h3>\r\nWith the [latex]{x}^{2}[\/latex] term isolated, the square root propty states that:\r\n<div style=\"text-align: center;\">[latex]\\text{if }{x}^{2}=k,\\text{then }x=\\pm \\sqrt{k}[\/latex]<\/div>\r\nwhere <em>k <\/em>is a nonzero real number.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: principal square root<\/h3>\r\nRecall that the\u00a0<em>princicpal square root<\/em> of a number such as [latex]\\sqrt{9}[\/latex] is the non-negative root, [latex]3[\/latex].\r\n\r\nAnd note that there is a difference between [latex]\\sqrt{9}[\/latex]\u00a0 and\u00a0[latex]{x}^{2}=9[\/latex]. In the case of\u00a0[latex]{x}^{2}=9[\/latex], we seek\u00a0<em>all numbers\u00a0<\/em>whose square is 9, that is\u00a0[latex]x=\\pm 3 [\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic equation with an [latex]{x}^{2}[\/latex] term but no [latex]x[\/latex] term, use the square root property to solve it<\/h3>\r\n<ol>\r\n \t<li>Isolate the [latex]{x}^{2}[\/latex] term on one side of the equal sign.<\/li>\r\n \t<li>Take the square root of both sides of the equation, putting a [latex]\\pm [\/latex] sign before the expression on the side opposite the squared term.<\/li>\r\n \t<li>Simplify the numbers on the side with the [latex]\\pm [\/latex] sign.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Simple Quadratic Equation Using the Square Root Property<\/h3>\r\nSolve the quadratic using the square root property: [latex]{x}^{2}=8[\/latex].\r\n\r\n[reveal-answer q=\"210107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"210107\"]\r\n\r\nTake the square root of both sides, and then simplify the radical. Remember to use a [latex]\\pm [\/latex] sign before the radical symbol.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}\\hfill&amp;=8\\hfill \\\\ x\\hfill&amp;=\\pm \\sqrt{8}\\hfill \\\\ \\hfill&amp;=\\pm 2\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=2\\sqrt{2}[\/latex], [latex]x=-2\\sqrt{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\r\nSolve the quadratic equation: [latex]4{x}^{2}+1=7[\/latex]\r\n\r\n[reveal-answer q=\"885054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"885054\"]\r\n\r\nFirst, isolate the [latex]{x}^{2}[\/latex] term. Then take the square root of both sides.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+1=7\\hfill \\\\ 4{x}^{2}=6\\hfill \\\\ {x}^{2}=\\frac{6}{4}\\hfill \\\\ x=\\pm \\frac{\\sqrt{6}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=\\frac{\\sqrt{6}}{2}[\/latex], [latex]x=-\\frac{\\sqrt{6}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation using the square root property: [latex]3{\\left(x - 4\\right)}^{2}=15[\/latex].\r\n\r\n[reveal-answer q=\"701664\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"701664\"]\r\n\r\n[latex]x=4\\pm \\sqrt{5}[\/latex]\r\n\r\n[latex]x=4\\sqrt{5}[\/latex],\u00a0[latex]-4\\sqrt{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]29172[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Completing the Square<\/h2>\r\nOne method is known as <strong>completing the square<\/strong>. Using this process, 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. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal 1. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n\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 sign 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}{l}\\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}{l}{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}{l}\\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 class=\"textbox examples\">\r\n<h3>Properties of Equality and taking the square root of both sides<\/h3>\r\nRemember that we are permitted, by the properties of equality, to add, subtract, multiply, or divide the same amount to both sides of an equation. Doing so won't change the value of the equation but it will enable us to isolate the variable on one side (that is, to solve the equation for the variable).\r\n\r\nThe square root property gives us another operation we can do to both sides of an equation, taking the square root. We just have to remember when taking the square root (or any even root, as we'll see later), to consider both the positive and negative possibilities of the constant.\r\n<p style=\"text-align: center;\">The square root property<\/p>\r\n<p style=\"text-align: center;\">If [latex]\\sqrt{x}=k[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Then [latex]x = \\pm{k}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic by Completing the Square<\/h3>\r\nSolve the quadratic equation by completing the square: [latex]{x}^{2}-3x - 5=0[\/latex].\r\n\r\n[reveal-answer q=\"676921\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676921\"]\r\n\r\nFirst, move the constant term to the right side of the equal sign by adding 5 to both sides of the equation.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\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<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve by completing the square: [latex]{x}^{2}-6x=13[\/latex].\r\n\r\n[reveal-answer q=\"222291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"222291\"]\r\n\r\n[latex]x=3\\pm \\sqrt{22}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nNote that when solving a quadratic by completing the square, a negative value will sometimes arise under the square root symbol. Later, we'll see that this value can be represented by a complex number (as shown in the video help for the problem below). We may also treat this type of solution as <em>unreal<\/em>, stating that no real solutions exist for this equation, by writing <em>DNE<\/em>. We will study complex numbers more thoroughly in a later module.\r\n\r\n[ohm_question]1384[\/ohm_question]\r\n\r\n[ohm_question]79619[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using the Quadratic Formula<\/h2>\r\nThe fourth method of solving a <strong>quadratic equation<\/strong> is by using the <strong>quadratic formula<\/strong>, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.\r\n\r\nWe can derive the quadratic formula by <strong>completing the square<\/strong>. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:\r\n<ol>\r\n \t<li>First, move the constant term to the right side of the equal sign:\r\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\r\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div><\/li>\r\n \t<li>Now, use the square root property, which gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Quadratic Formula<\/h3>\r\nWritten in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], any quadratic equation can be solved using the <strong>quadratic formula<\/strong>:\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\r\nwhere <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic equation, solve it using the quadratic formula<\/h3>\r\n<ol>\r\n \t<li>Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[\/latex].<\/li>\r\n \t<li>Make note of the values of the coefficients and constant term, [latex]a,b[\/latex], and [latex]c[\/latex].<\/li>\r\n \t<li>Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\r\n \t<li>Calculate and solve.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example : Solve A Quadratic Equation Using the Quadratic Formula<\/h3>\r\nSolve the quadratic equation: [latex]{x}^{2}+5x+1=0[\/latex].\r\n\r\n[reveal-answer q=\"641400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"641400\"]\r\n\r\nIdentify the coefficients: [latex]a=1,b=5,c=1[\/latex]. Then use the quadratic formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&amp;=\\frac{-\\left(5\\right)\\pm \\sqrt{{\\left(5\\right)}^{2}-4\\left(1\\right)\\left(1\\right)}}{2\\left(1\\right)}\\hfill \\\\ \\hfill&amp;=\\frac{-5\\pm \\sqrt{25 - 4}}{2}\\hfill \\\\ \\hfill&amp;=\\frac{-5\\pm \\sqrt{21}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall adding and subtracting fractions<\/h3>\r\nThe form we have used to recall adding and subtracting fractions can help us write the solutions to quadratic equations.\r\n\r\n[latex]\\dfrac{a}{b}\\pm\\dfrac{c}{d} = \\dfrac{ad \\pm bc}{bd}[\/latex]\r\n\r\nEx. The solutions\r\n\r\n[latex]x=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]\r\n\r\ncan also be written as two separate fractions\r\n\r\n[latex]x=\\dfrac{-b}{2a} \\pm \\dfrac{\\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\r\nUse the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].\r\n\r\n[reveal-answer q=\"688902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688902\"]\r\n\r\nFirst, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex].\r\n\r\nSubstitute these values into the quadratic formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&amp;=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&amp;=\\dfrac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&amp;=\\dfrac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&amp;=\\dfrac{-1\\pm \\sqrt{-7}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\nThe solutions to the equation are <em>unreal <\/em>because the square root of a negative number does not exist in the real numbers. If asked to find all <em>real<\/em> solutions, we would indicate that they do not exist by writing\u00a0<em>DNE<\/em>.\r\n\r\nWe may also express unreal solutions as complex numbers by writing\r\n\r\n[latex]x=\\dfrac{-1\\pm i\\sqrt{7}}{2}[\/latex]\r\n\r\nor\r\n\r\n[latex]x=-\\dfrac{1}{2}+\\dfrac{\\sqrt{7}}{2}i[\/latex] and [latex]x=-\\dfrac{1}{2}-\\dfrac{\\sqrt{7}}{2}i[\/latex]\u00a0,\u00a0where [latex]\\sqrt{-1}[\/latex] is expressed as [latex]i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[\/latex].\r\n\r\n[reveal-answer q=\"232269\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"232269\"]\r\n\r\n[latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex][\/hidden-answer]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4014&amp;theme=oea&amp;iframe_resize_id=mom3[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35639&amp;theme=oea&amp;iframe_resize_id=mom4[\/embed]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions.<\/li>\r\n \t<li>Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method.<\/li>\r\n \t<li>Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution.<\/li>\r\n \t<li>Completing the square is a method of solving quadratic equations when the equation cannot be factored.<\/li>\r\n \t<li>A highly dependable method for solving quadratic equations is the quadratic formula based on the coefficients and the constant term in the equation.<\/li>\r\n \t<li>The discriminant is used to indicate the nature of the solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.<\/li>\r\n \t<li>The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165131990658\" class=\"definition\">\r\n \t<dt><strong>completing the square<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165131990661\">a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943522\" class=\"definition\">\r\n \t<dt><strong>discriminant<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165132943525\">the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943528\" class=\"definition\">\r\n \t<dt><strong>Pythagorean Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134297639\">a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165131990658\" class=\"definition\">\r\n \t<dt><strong>quadratic equation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165131990661\">an equation containing a second-degree polynomial; can be solved using multiple methods<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943522\" class=\"definition\">\r\n \t<dt><strong>quadratic formula<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165132943525\">a formula that will solve all quadratic equations<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943528\" class=\"definition\">\r\n \t<dt><strong>square root property<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134297639\">one of the methods used to solve a quadratic equation in which the [latex]{x}^{2}[\/latex] term is isolated so that the square root of both sides of the equation can be taken to solve for <em>x<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943528\" class=\"definition\">\r\n \t<dt><strong>zero-product property<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134297639\">the property that formally states that multiplication by zero is zero so that each factor of a quadratic equation can be set equal to zero to solve equations<\/dd>\r\n<\/dl>\r\n&nbsp;\r\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Section CR 14 Homework Exercises<\/span><\/h2>\r\n1. How do we recognize when an equation is quadratic?\r\n\r\n2. When we solve a quadratic equation, how many solutions should we always start seeking out? Explain why when solving a quadratic equation in the form [latex]ax^{2}+bx+c=0[\/latex] we may graph the equation [latex]\\text{ }y=ax^{2}+bx+c[\/latex] and have no zeroes ([latex]x[\/latex]-intercepts).\r\n\r\n3. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?\r\n\r\n4. In the quadratic formula, what is the name of the expression under the radical sign [latex]\\text{ }b^{2}-4ac[\/latex], and how does it determine the number of and nature of our solutions?\r\n\r\n5. Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.\r\n\r\nFor the following exercises, solve the quadratic equation by factoring.\r\n\r\n6. [latex]\\text{ }x^{2}-4x-21=0[\/latex]\r\n\r\n7. [latex]\\text{ }x^{2}-9x+18=0[\/latex]\r\n\r\n8. [latex]\\text{ }2x^{2}+9x-5=0[\/latex]\r\n\r\n9. [latex]\\text{ }6x^{2}+17x+5=0[\/latex]\r\n\r\n10. [latex]\\text{ }4x^{2}-12x+8=0[\/latex]\r\n\r\n11. [latex]\\text{ }3x^{2}-75=0[\/latex]\r\n\r\n12. [latex]\\text{ }8x^{2}+6x-9=0[\/latex]\r\n\r\n13. [latex]\\text{ }4x^{2}=9[\/latex]\r\n\r\n14. [latex]\\text{ }2x^{2}+14x=36[\/latex]\r\n\r\n15. [latex]\\text{ }5x^{2}=5x+30[\/latex]\r\n\r\n16. [latex]\\text{ }4x^{2}=5x[\/latex]\r\n\r\n17. [latex]\\text{ }7x^{2}+3x=0[\/latex]\r\n\r\n18. [latex]\\text{ }\\dfrac{x}{3}-\\dfrac{9}{x}=2[\/latex]\r\n\r\nFor the following exercises, solve the quadratic equation by using the square root property.\r\n\r\n19. [latex]\\text{ }x^{2}=36[\/latex]\r\n\r\n20. [latex]\\text{ }x^{2}=49[\/latex]\r\n\r\n21. [latex]\\text{ }(x-1)^{2}=25[\/latex]\r\n\r\n22. [latex]\\text{ }(x-3)^{2}=7[\/latex]\r\n\r\n23. [latex]\\text{ }(2x+1)^{2}=9[\/latex]\r\n\r\n24. [latex]\\text{ }(x-5)^{2}=4[\/latex]\r\n\r\nFor the following exercises, solve the quadratic equation by completing the square. Show each step.\r\n\r\n25. [latex]\\text{ }x^{2}-9x-22=0[\/latex]\r\n\r\n26. [latex]\\text{ }2x^{2}-8x-5=0[\/latex]\r\n\r\n27. [latex]\\text{ }x^{2}-6x=13[\/latex]\r\n\r\n28. [latex]\\text{ }x^{2}+\\dfrac{2}{3}x-\\dfrac{1}{3}=0[\/latex]\r\n\r\n29. [latex]\\text{ }2+z=6z^{2}[\/latex]\r\n\r\n30. [latex]\\text{ }6p^{2}+7p-20=0[\/latex]\r\n\r\n31. [latex]\\text{ }2x^{2}-3x-1=0[\/latex]\r\n\r\nFor the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.\r\n\r\n32. [latex]\\text{ }2x^{2}-6x+7=0[\/latex]\r\n\r\n33. [latex]\\text{ }x^{2}+4x+7=0[\/latex]\r\n\r\n34. [latex]\\text{ }3x^{2}+5x-8=0[\/latex]\r\n\r\n35. [latex]\\text{ }9x^{2}-30x+25=0[\/latex]\r\n\r\n36. [latex]\\text{ }2x^{2}-3x-7=0[\/latex]\r\n\r\n37. [latex]\\text{ }6x^{2}-x-2=0[\/latex]\r\n\r\nFor the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state <i>No Real Solution<\/i>.\r\n\r\n38. [latex]\\text{ }2x^{2}+5x+3=0[\/latex]\r\n\r\n39. [latex]\\text{ }x^{2}+x=4[\/latex]\r\n\r\n40. [latex]\\text{ }2x^{2}-8x-5=0[\/latex]\r\n\r\n41. [latex]\\text{ }3x^{2}-5x+1=0[\/latex]\r\n\r\n42. [latex]\\text{ }x^{2}+4x+2=0[\/latex]\r\n\r\n43. [latex]\\text{ }4+\\dfrac{1}{x}-\\dfrac{1}{x^{2}}=0[\/latex]\r\n\r\n44. [latex]\\text{ }x+\\dfrac{4}{x}+6=0[\/latex]\r\n\r\n45. Beginning with the general form of the quadratic equation [latex]\\text{ }ax^{2}+bx+c=0[\/latex], solve for x by using the completing the square method, thus deriving the quadratic formula.\r\n\r\n46. Show that the sum of the two solutions to the quadratic formula is [latex]-\\dfrac{b}{a}[\/latex].\r\n\r\n47. A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is [latex]119 ft^{2}[\/latex]. Solve the quadratic equation to find the length and width.\r\n\r\n48. Abercrombie and Fitch stock had a price given as [latex]P=0.2t^{2}-5.6t+50.2[\/latex], where [latex]t[\/latex] is the time in months from 1999 to 2001. ([latex]t=1[\/latex] is January 1999). Find the two months in which the price of the stock was $30.\r\n\r\n49. Suppose that an equation is given [latex]p=-2x^{2}+280x-1000[\/latex], where [latex]x[\/latex] represents the number of items sold at an auction and [latex]p[\/latex] is the profit made by the business that ran the auction. How many items solve would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using <b>2nd CALC maximum<\/b>. To obtain a good window for the curve, set [latex]x[\/latex] [0,200] and [latex]y[\/latex] [0,10000].\r\n\r\n50. A formula for the normal systolic blood pressure for a man age A, measured in mmHg, is given as [latex]P=0.006A^{2}-0.02A+120[\/latex]. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.\r\n\r\n51. The cost function for a certain company is [latex]C=60x+300[\/latex] and the revenue is given by [latex]R=100x-0.5x^{2}[\/latex]. Recall the the profit is revenue minus cost. Set up a quadratic equation and find two values of [latex]x[\/latex] (production level) that will create a profit of $300.\r\n\r\n52. A falling object travels a distance given by the formula [latex]d=5t+16t^{2}[\/latex] ft, where [latex]t[\/latex] is measured in seconds. How long will it take for the object to travel 74 ft?\r\n\r\n53. A vacanot lot is being converted into a community garden. The garden and walkway around its perimeter have an area of [latex]378\\text{ }ft^{2}[\/latex]. Find the width of the walkway if the garden is [latex]12ft[\/latex] wide by [latex]15ft[\/latex] long.\r\n<img src=\"http:\/\/www.hutchmath.com\/Images\/Sec3_1Prob_57.JPG\" alt=\"Garden with walkway border.\" \/>\r\n\r\n54. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu [latex]t[\/latex] days after it broke out is given by the model [latex]P=-t^{2}+13t+130[\/latex], where [latex]1&lt;=t&lt;=6[\/latex]. Find the day that 160 students had the flu. Recall that the restriction on [latex]t[\/latex] is at most 6.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a quadratic equation to solve it.<\/li>\n<li>Use the square root property to solve a quadratic equation.<\/li>\n<li>Complete the square to solve a quadratic equation.<\/li>\n<li>Use the quadratic formula to solve a quadratic equation.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: identifying polynomials<\/h3>\n<p>Recall that a polynomial is an expression containing variable terms joined together by addition or subtraction. We can identify the\u00a0<strong>degree<\/strong> of the polynomial by finding the term having the highest power (largest exponent on its variable). That term will be the\u00a0<strong>leading term<\/strong> of the polynomial and its degree is the degree of the polynomial.<\/p>\n<\/div>\n<p>An equation containing a second-degree polynomial is called a <strong>quadratic equation<\/strong>. For example, equations such as [latex]2{x}^{2}+3x - 1=0[\/latex] and [latex]{x}^{2}-4=0[\/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.<\/p>\n<p>Often the easiest method of solving a quadratic equation is <strong>factoring<\/strong>. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.<\/p>\n<p>If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if [latex]a\\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where <em>a <\/em>and <em>b <\/em>are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.<\/p>\n<p>Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression [latex]\\left(x - 2\\right)\\left(x+3\\right)[\/latex] by multiplying the two factors together.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)\\hfill&={x}^{2}+3x - 2x - 6\\hfill \\\\ \\hfill&={x}^{2}+x - 6\\hfill \\end{array}[\/latex]<\/div>\n<p>The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[\/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.<\/p>\n<p>The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], where <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex]. The equation [latex]{x}^{2}+x - 6=0[\/latex] is in standard form.<\/p>\n<div class=\"textbox examples\">\n<h3>recall the greatest common factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) is the largest number or variable expression that can divided evenly into two or more numbers or expressions.<\/p>\n<p>To find the GCF of an expression containing variable terms, first find the GCF of the coefficients, then find the GCF of the variables. The GCF of the variables will be the smallest degree of each variable that appears in each term.<\/p>\n<p>For example,\u00a0[latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest expression that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\n<p>When factoring, remember to look first for a greatest common factor. If you can find one to factor out, it will usually make your factoring task less challenging.<\/p>\n<\/div>\n<p>We can use the zero-product property to solve quadratic equations in which we first have to factor out the <strong>greatest common factor<\/strong> (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Zero-Product Property and Quadratic Equations<\/h3>\n<p>The <strong>zero-product property<\/strong> states<\/p>\n<div style=\"text-align: center;\">[latex]\\text{If }a\\cdot b=0,\\text{ then }a=0\\text{ or }b=0[\/latex],<\/div>\n<p>where <em>a <\/em>and <em>b <\/em>are real numbers or algebraic expressions.<\/p>\n<p>A <strong>quadratic equation<\/strong> is an equation containing a second-degree polynomial; for example<\/p>\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx+c=0[\/latex]<\/div>\n<p>where <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers, and [latex]a\\ne 0[\/latex]. It is in standard form.<\/p>\n<\/div>\n<h2>Solving Quadratics with a Leading Coefficient of 1<\/h2>\n<p>In the quadratic equation [latex]{x}^{2}+x - 6=0[\/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[\/latex], is 1. We have one method of factoring quadratic equations in this form.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic equation with the leading coefficient of 1, factor it<\/h3>\n<ol>\n<li>Find two numbers whose product equals <em>c<\/em> and whose sum equals <em>b<\/em>.<\/li>\n<li>Use those numbers to write two factors of the form [latex]\\left(x+k\\right)\\text{ or }\\left(x-k\\right)[\/latex], where <em>k <\/em>is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and [latex]-2[\/latex], the factors are [latex]\\left(x+1\\right)\\left(x - 2\\right)[\/latex].<\/li>\n<li>Solve using the zero-product property by setting each factor equal to zero and solving for the variable.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring and Solving a Quadratic with Leading Coefficient of 1<\/h3>\n<p>Factor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16111\">Show Solution<\/span><\/p>\n<div id=\"q16111\" class=\"hidden-answer\" style=\"display: none\">\n<p>To factor [latex]{x}^{2}+x - 6=0[\/latex], we look for two numbers whose product equals [latex]-6[\/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}1\\cdot \\left(-6\\right)\\hfill \\\\ \\left(-6\\right)\\cdot 1\\hfill \\\\ 2\\cdot \\left(-3\\right)\\hfill \\\\ 3\\cdot \\left(-2\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>The last pair, [latex]3\\cdot \\left(-2\\right)[\/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.<\/p>\n<div style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/div>\n<p>To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/div>\n<p>The two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/latex]. We can see how the solutions relate to the graph below. The solutions are the <em>x-<\/em>intercepts of the graph of [latex]{x}^{2}+x - 6[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24224527\/CNX_CAT_Figure_02_05_002.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well.\" width=\"487\" height=\"588\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor and solve the quadratic equation: [latex]{x}^{2}-5x - 6=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q220537\">Show Solution<\/span><\/p>\n<div id=\"q220537\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(x - 6\\right)\\left(x+1\\right)=0;x=6,x=-1[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm2029\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2029&theme=oea&iframe_resize_id=ohm2029&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using Other Factoring Methods<\/h2>\n<div class=\"textbox exercises\">\n<h3>Example: Setting it Equal to Zero and Factoring<\/h3>\n<p>Factor and solve the equation: [latex]{x}^{3}=100{x}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16120\">Show Solution<\/span><\/p>\n<div id=\"q16120\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we must get everything equal to zero. If we divide both sides by x instead, we may miss one of our answers. That is why we must first set the equation equal to zero:<br \/>\n[latex]{x}^{3}-100{x}^{2}=0[\/latex]<\/p>\n<p>Next we will factor out the GCF: [latex]{x}^{2}(x-100)=0[\/latex].<\/p>\n<p>To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {x}^{2}(x-100)=0\\hfill \\\\x^2=0 \\hfill\\\\ x=0\\hfill \\\\ (x - 100)=0\\hfill \\\\ x=100\\hfill \\end{array}[\/latex]<\/div>\n<p>The two solutions are [latex]x=0[\/latex] and [latex]x=100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor and solve the quadratic equation: [latex]{x}^{20}=2{x}^{19}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q220540\">Show Solution<\/span><\/p>\n<div id=\"q220540\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{x}^{19}\\left(x - 2\\right)=0;x=0,x=2[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm248381\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=248381&theme=oea&iframe_resize_id=ohm248381&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Method of Factor by Grouping<\/h3>\n<p>Factor and solve the equation: [latex]{x}^{3}+11{x}^{2}-121x-1331=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16130\">Show Solution<\/span><\/p>\n<div id=\"q16130\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will solve this we will use the the method of factoring by grouping. First group the terms.<br \/>\n[latex]{x}^{3}+11{x}^{2}-121x-1331=0[\/latex]<br \/>\n[latex]({x}^{3}+11{x}^{2})-(121x+1331)=0[\/latex]<\/p>\n<p>The greatest common factor in the first grouping is [latex]{x}^{2}[\/latex].<br \/>\nThe greatest common factor in the second grouping is [latex]121[\/latex].<\/p>\n<p>Factor out [latex]{x}^{2}[\/latex] from the first grouping and [latex]121[\/latex] from the second grouping.<br \/>\n[latex]({x}^{3}+11{x}^{2})-(121x+1331)=0[\/latex]<br \/>\n[latex]{x}^{2}(x+11)-121(x+11)=0[\/latex]<\/p>\n<p>The common factor in the [latex]{x}^{2}(x+11)-121(x+11)=0[\/latex] is [latex](x+11)[\/latex]. We will now factor this out:<br \/>\n[latex](x+11)({x}^{2}-121)=0[\/latex]<\/p>\n<p>We can factor one more time:<br \/>\n[latex](x+11)(x+11)(x-11)=0[\/latex]<\/p>\n<p>To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} (x+11)(x+11)(x-11)=0\\hfill \\\\(x+11)=0 \\hfill\\\\ x=-11\\hfill \\\\ (x +11)=0\\hfill \\\\ x=-11\\hfill \\\\(x-11)=0 \\hfill\\\\ x=11\\hfill \\end{array}[\/latex]<\/div>\n<p>One solution is repeated twice, but we only need to write it once.<br \/>\nTherefore, the two solutions are [latex]x=-11[\/latex] and [latex]x=11[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor and solve the quadratic equation: [latex]{x}^{3}-12{x}^2-144x+1728=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q220550\">Show Solution<\/span><\/p>\n<div id=\"q220550\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{x}^{3}-12{x}^2-144x+1728=0;x=-12,x=12[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm248391\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=248391&theme=oea&iframe_resize_id=ohm248391&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Square Root Property<\/h2>\n<p>When there is no linear term in the equation, another method of solving a quadratic equation is by using the <strong>square root property<\/strong>, in which we isolate the [latex]{x}^{2}[\/latex] term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the [latex]{x}^{2}[\/latex] term so that the square root property can be used.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Square Root Property<\/h3>\n<p>With the [latex]{x}^{2}[\/latex] term isolated, the square root propty states that:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{if }{x}^{2}=k,\\text{then }x=\\pm \\sqrt{k}[\/latex]<\/div>\n<p>where <em>k <\/em>is a nonzero real number.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: principal square root<\/h3>\n<p>Recall that the\u00a0<em>princicpal square root<\/em> of a number such as [latex]\\sqrt{9}[\/latex] is the non-negative root, [latex]3[\/latex].<\/p>\n<p>And note that there is a difference between [latex]\\sqrt{9}[\/latex]\u00a0 and\u00a0[latex]{x}^{2}=9[\/latex]. In the case of\u00a0[latex]{x}^{2}=9[\/latex], we seek\u00a0<em>all numbers\u00a0<\/em>whose square is 9, that is\u00a0[latex]x=\\pm 3[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic equation with an [latex]{x}^{2}[\/latex] term but no [latex]x[\/latex] term, use the square root property to solve it<\/h3>\n<ol>\n<li>Isolate the [latex]{x}^{2}[\/latex] term on one side of the equal sign.<\/li>\n<li>Take the square root of both sides of the equation, putting a [latex]\\pm[\/latex] sign before the expression on the side opposite the squared term.<\/li>\n<li>Simplify the numbers on the side with the [latex]\\pm[\/latex] sign.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Simple Quadratic Equation Using the Square Root Property<\/h3>\n<p>Solve the quadratic using the square root property: [latex]{x}^{2}=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q210107\">Show Solution<\/span><\/p>\n<div id=\"q210107\" class=\"hidden-answer\" style=\"display: none\">\n<p>Take the square root of both sides, and then simplify the radical. Remember to use a [latex]\\pm[\/latex] sign before the radical symbol.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}\\hfill&=8\\hfill \\\\ x\\hfill&=\\pm \\sqrt{8}\\hfill \\\\ \\hfill&=\\pm 2\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=2\\sqrt{2}[\/latex], [latex]x=-2\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\n<p>Solve the quadratic equation: [latex]4{x}^{2}+1=7[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q885054\">Show Solution<\/span><\/p>\n<div id=\"q885054\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, isolate the [latex]{x}^{2}[\/latex] term. Then take the square root of both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+1=7\\hfill \\\\ 4{x}^{2}=6\\hfill \\\\ {x}^{2}=\\frac{6}{4}\\hfill \\\\ x=\\pm \\frac{\\sqrt{6}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=\\frac{\\sqrt{6}}{2}[\/latex], [latex]x=-\\frac{\\sqrt{6}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation using the square root property: [latex]3{\\left(x - 4\\right)}^{2}=15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q701664\">Show Solution<\/span><\/p>\n<div id=\"q701664\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=4\\pm \\sqrt{5}[\/latex]<\/p>\n<p>[latex]x=4\\sqrt{5}[\/latex],\u00a0[latex]-4\\sqrt{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm29172\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29172&theme=oea&iframe_resize_id=ohm29172&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Completing the Square<\/h2>\n<p>One method is known as <strong>completing the square<\/strong>. Using this process, 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. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal 1. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\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 sign 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}{l}\\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}{l}{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}{l}\\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 class=\"textbox examples\">\n<h3>Properties of Equality and taking the square root of both sides<\/h3>\n<p>Remember that we are permitted, by the properties of equality, to add, subtract, multiply, or divide the same amount to both sides of an equation. Doing so won&#8217;t change the value of the equation but it will enable us to isolate the variable on one side (that is, to solve the equation for the variable).<\/p>\n<p>The square root property gives us another operation we can do to both sides of an equation, taking the square root. We just have to remember when taking the square root (or any even root, as we&#8217;ll see later), to consider both the positive and negative possibilities of the constant.<\/p>\n<p style=\"text-align: center;\">The square root property<\/p>\n<p style=\"text-align: center;\">If [latex]\\sqrt{x}=k[\/latex]<\/p>\n<p style=\"text-align: center;\">Then [latex]x = \\pm{k}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic by Completing the Square<\/h3>\n<p>Solve the quadratic equation 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=\"q676921\">Show Solution<\/span><\/p>\n<div id=\"q676921\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, move the constant term to the right side of the equal sign by adding 5 to both sides of the equation.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\n<p>Then, 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<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve by completing the square: [latex]{x}^{2}-6x=13[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q222291\">Show Solution<\/span><\/p>\n<div id=\"q222291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3\\pm \\sqrt{22}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Note that when solving a quadratic by completing the square, a negative value will sometimes arise under the square root symbol. Later, we&#8217;ll see that this value can be represented by a complex number (as shown in the video help for the problem below). We may also treat this type of solution as <em>unreal<\/em>, stating that no real solutions exist for this equation, by writing <em>DNE<\/em>. We will study complex numbers more thoroughly in a later module.<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm1384\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1384&theme=oea&iframe_resize_id=ohm1384&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm79619\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79619&theme=oea&iframe_resize_id=ohm79619&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Quadratic Formula<\/h2>\n<p>The fourth method of solving a <strong>quadratic equation<\/strong> is by using the <strong>quadratic formula<\/strong>, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.<\/p>\n<p>We can derive the quadratic formula by <strong>completing the square<\/strong>. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:<\/p>\n<ol>\n<li>First, move the constant term to the right side of the equal sign:\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div>\n<\/li>\n<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div>\n<\/li>\n<li>Now, use the square root property, which gives\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<div class=\"textbox\">\n<h3>A General Note: The Quadratic Formula<\/h3>\n<p>Written in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], any quadratic equation can be solved using the <strong>quadratic formula<\/strong>:<\/p>\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<p>where <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic equation, solve it using the quadratic formula<\/h3>\n<ol>\n<li>Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[\/latex].<\/li>\n<li>Make note of the values of the coefficients and constant term, [latex]a,b[\/latex], and [latex]c[\/latex].<\/li>\n<li>Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\n<li>Calculate and solve.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example : Solve A Quadratic Equation Using the Quadratic Formula<\/h3>\n<p>Solve the quadratic equation: [latex]{x}^{2}+5x+1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q641400\">Show Solution<\/span><\/p>\n<div id=\"q641400\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify the coefficients: [latex]a=1,b=5,c=1[\/latex]. Then use the quadratic formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&=\\frac{-\\left(5\\right)\\pm \\sqrt{{\\left(5\\right)}^{2}-4\\left(1\\right)\\left(1\\right)}}{2\\left(1\\right)}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{25 - 4}}{2}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{21}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall adding and subtracting fractions<\/h3>\n<p>The form we have used to recall adding and subtracting fractions can help us write the solutions to quadratic equations.<\/p>\n<p>[latex]\\dfrac{a}{b}\\pm\\dfrac{c}{d} = \\dfrac{ad \\pm bc}{bd}[\/latex]<\/p>\n<p>Ex. The solutions<\/p>\n<p>[latex]x=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/p>\n<p>can also be written as two separate fractions<\/p>\n<p>[latex]x=\\dfrac{-b}{2a} \\pm \\dfrac{\\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\n<p>Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688902\">Show Solution<\/span><\/p>\n<div id=\"q688902\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex].<\/p>\n<p>Substitute these values into the quadratic formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&=\\dfrac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&=\\dfrac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&=\\dfrac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&=\\dfrac{-1\\pm \\sqrt{-7}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<p>The solutions to the equation are <em>unreal <\/em>because the square root of a negative number does not exist in the real numbers. If asked to find all <em>real<\/em> solutions, we would indicate that they do not exist by writing\u00a0<em>DNE<\/em>.<\/p>\n<p>We may also express unreal solutions as complex numbers by writing<\/p>\n<p>[latex]x=\\dfrac{-1\\pm i\\sqrt{7}}{2}[\/latex]<\/p>\n<p>or<\/p>\n<p>[latex]x=-\\dfrac{1}{2}+\\dfrac{\\sqrt{7}}{2}i[\/latex] and [latex]x=-\\dfrac{1}{2}-\\dfrac{\\sqrt{7}}{2}i[\/latex]\u00a0,\u00a0where [latex]\\sqrt{-1}[\/latex] is expressed as [latex]i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q232269\">Show Solution<\/span><\/p>\n<div id=\"q232269\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm4014\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4014&#38;theme=oea&#38;iframe_resize_id=ohm4014&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm35639\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35639&#38;theme=oea&#38;iframe_resize_id=ohm35639&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions.<\/li>\n<li>Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method.<\/li>\n<li>Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution.<\/li>\n<li>Completing the square is a method of solving quadratic equations when the equation cannot be factored.<\/li>\n<li>A highly dependable method for solving quadratic equations is the quadratic formula based on the coefficients and the constant term in the equation.<\/li>\n<li>The discriminant is used to indicate the nature of the solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.<\/li>\n<li>The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165131990658\" class=\"definition\">\n<dt><strong>completing the square<\/strong><\/dt>\n<dd id=\"fs-id1165131990661\">a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943522\" class=\"definition\">\n<dt><strong>discriminant<\/strong><\/dt>\n<dd id=\"fs-id1165132943525\">the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943528\" class=\"definition\">\n<dt><strong>Pythagorean Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165134297639\">a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems<\/dd>\n<\/dl>\n<dl id=\"fs-id1165131990658\" class=\"definition\">\n<dt><strong>quadratic equation<\/strong><\/dt>\n<dd id=\"fs-id1165131990661\">an equation containing a second-degree polynomial; can be solved using multiple methods<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943522\" class=\"definition\">\n<dt><strong>quadratic formula<\/strong><\/dt>\n<dd id=\"fs-id1165132943525\">a formula that will solve all quadratic equations<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943528\" class=\"definition\">\n<dt><strong>square root property<\/strong><\/dt>\n<dd id=\"fs-id1165134297639\">one of the methods used to solve a quadratic equation in which the [latex]{x}^{2}[\/latex] term is isolated so that the square root of both sides of the equation can be taken to solve for <em>x<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943528\" class=\"definition\">\n<dt><strong>zero-product property<\/strong><\/dt>\n<dd id=\"fs-id1165134297639\">the property that formally states that multiplication by zero is zero so that each factor of a quadratic equation can be set equal to zero to solve equations<\/dd>\n<\/dl>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\"><span style=\"text-decoration: underline;\">Section CR 14 Homework Exercises<\/span><\/h2>\n<p>1. How do we recognize when an equation is quadratic?<\/p>\n<p>2. When we solve a quadratic equation, how many solutions should we always start seeking out? Explain why when solving a quadratic equation in the form [latex]ax^{2}+bx+c=0[\/latex] we may graph the equation [latex]\\text{ }y=ax^{2}+bx+c[\/latex] and have no zeroes ([latex]x[\/latex]-intercepts).<\/p>\n<p>3. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?<\/p>\n<p>4. In the quadratic formula, what is the name of the expression under the radical sign [latex]\\text{ }b^{2}-4ac[\/latex], and how does it determine the number of and nature of our solutions?<\/p>\n<p>5. Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.<\/p>\n<p>For the following exercises, solve the quadratic equation by factoring.<\/p>\n<p>6. [latex]\\text{ }x^{2}-4x-21=0[\/latex]<\/p>\n<p>7. [latex]\\text{ }x^{2}-9x+18=0[\/latex]<\/p>\n<p>8. [latex]\\text{ }2x^{2}+9x-5=0[\/latex]<\/p>\n<p>9. [latex]\\text{ }6x^{2}+17x+5=0[\/latex]<\/p>\n<p>10. [latex]\\text{ }4x^{2}-12x+8=0[\/latex]<\/p>\n<p>11. [latex]\\text{ }3x^{2}-75=0[\/latex]<\/p>\n<p>12. [latex]\\text{ }8x^{2}+6x-9=0[\/latex]<\/p>\n<p>13. [latex]\\text{ }4x^{2}=9[\/latex]<\/p>\n<p>14. [latex]\\text{ }2x^{2}+14x=36[\/latex]<\/p>\n<p>15. [latex]\\text{ }5x^{2}=5x+30[\/latex]<\/p>\n<p>16. [latex]\\text{ }4x^{2}=5x[\/latex]<\/p>\n<p>17. [latex]\\text{ }7x^{2}+3x=0[\/latex]<\/p>\n<p>18. [latex]\\text{ }\\dfrac{x}{3}-\\dfrac{9}{x}=2[\/latex]<\/p>\n<p>For the following exercises, solve the quadratic equation by using the square root property.<\/p>\n<p>19. [latex]\\text{ }x^{2}=36[\/latex]<\/p>\n<p>20. [latex]\\text{ }x^{2}=49[\/latex]<\/p>\n<p>21. [latex]\\text{ }(x-1)^{2}=25[\/latex]<\/p>\n<p>22. [latex]\\text{ }(x-3)^{2}=7[\/latex]<\/p>\n<p>23. [latex]\\text{ }(2x+1)^{2}=9[\/latex]<\/p>\n<p>24. [latex]\\text{ }(x-5)^{2}=4[\/latex]<\/p>\n<p>For the following exercises, solve the quadratic equation by completing the square. Show each step.<\/p>\n<p>25. [latex]\\text{ }x^{2}-9x-22=0[\/latex]<\/p>\n<p>26. [latex]\\text{ }2x^{2}-8x-5=0[\/latex]<\/p>\n<p>27. [latex]\\text{ }x^{2}-6x=13[\/latex]<\/p>\n<p>28. [latex]\\text{ }x^{2}+\\dfrac{2}{3}x-\\dfrac{1}{3}=0[\/latex]<\/p>\n<p>29. [latex]\\text{ }2+z=6z^{2}[\/latex]<\/p>\n<p>30. [latex]\\text{ }6p^{2}+7p-20=0[\/latex]<\/p>\n<p>31. [latex]\\text{ }2x^{2}-3x-1=0[\/latex]<\/p>\n<p>For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.<\/p>\n<p>32. [latex]\\text{ }2x^{2}-6x+7=0[\/latex]<\/p>\n<p>33. [latex]\\text{ }x^{2}+4x+7=0[\/latex]<\/p>\n<p>34. [latex]\\text{ }3x^{2}+5x-8=0[\/latex]<\/p>\n<p>35. [latex]\\text{ }9x^{2}-30x+25=0[\/latex]<\/p>\n<p>36. [latex]\\text{ }2x^{2}-3x-7=0[\/latex]<\/p>\n<p>37. [latex]\\text{ }6x^{2}-x-2=0[\/latex]<\/p>\n<p>For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state <i>No Real Solution<\/i>.<\/p>\n<p>38. [latex]\\text{ }2x^{2}+5x+3=0[\/latex]<\/p>\n<p>39. [latex]\\text{ }x^{2}+x=4[\/latex]<\/p>\n<p>40. [latex]\\text{ }2x^{2}-8x-5=0[\/latex]<\/p>\n<p>41. [latex]\\text{ }3x^{2}-5x+1=0[\/latex]<\/p>\n<p>42. [latex]\\text{ }x^{2}+4x+2=0[\/latex]<\/p>\n<p>43. [latex]\\text{ }4+\\dfrac{1}{x}-\\dfrac{1}{x^{2}}=0[\/latex]<\/p>\n<p>44. [latex]\\text{ }x+\\dfrac{4}{x}+6=0[\/latex]<\/p>\n<p>45. Beginning with the general form of the quadratic equation [latex]\\text{ }ax^{2}+bx+c=0[\/latex], solve for x by using the completing the square method, thus deriving the quadratic formula.<\/p>\n<p>46. Show that the sum of the two solutions to the quadratic formula is [latex]-\\dfrac{b}{a}[\/latex].<\/p>\n<p>47. A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is [latex]119 ft^{2}[\/latex]. Solve the quadratic equation to find the length and width.<\/p>\n<p>48. Abercrombie and Fitch stock had a price given as [latex]P=0.2t^{2}-5.6t+50.2[\/latex], where [latex]t[\/latex] is the time in months from 1999 to 2001. ([latex]t=1[\/latex] is January 1999). Find the two months in which the price of the stock was $30.<\/p>\n<p>49. Suppose that an equation is given [latex]p=-2x^{2}+280x-1000[\/latex], where [latex]x[\/latex] represents the number of items sold at an auction and [latex]p[\/latex] is the profit made by the business that ran the auction. How many items solve would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using <b>2nd CALC maximum<\/b>. To obtain a good window for the curve, set [latex]x[\/latex] [0,200] and [latex]y[\/latex] [0,10000].<\/p>\n<p>50. A formula for the normal systolic blood pressure for a man age A, measured in mmHg, is given as [latex]P=0.006A^{2}-0.02A+120[\/latex]. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.<\/p>\n<p>51. The cost function for a certain company is [latex]C=60x+300[\/latex] and the revenue is given by [latex]R=100x-0.5x^{2}[\/latex]. Recall the the profit is revenue minus cost. Set up a quadratic equation and find two values of [latex]x[\/latex] (production level) that will create a profit of $300.<\/p>\n<p>52. A falling object travels a distance given by the formula [latex]d=5t+16t^{2}[\/latex] ft, where [latex]t[\/latex] is measured in seconds. How long will it take for the object to travel 74 ft?<\/p>\n<p>53. A vacanot lot is being converted into a community garden. The garden and walkway around its perimeter have an area of [latex]378\\text{ }ft^{2}[\/latex]. Find the width of the walkway if the garden is [latex]12ft[\/latex] wide by [latex]15ft[\/latex] long.<br \/>\n<img decoding=\"async\" src=\"http:\/\/www.hutchmath.com\/Images\/Sec3_1Prob_57.JPG\" alt=\"Garden with walkway border.\" \/><\/p>\n<p>54. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu [latex]t[\/latex] days after it broke out is given by the model [latex]P=-t^{2}+13t+130[\/latex], where [latex]1<=t<=6[\/latex]. Find the day that 160 students had the flu. Recall that the restriction on [latex]t[\/latex] is at most 6.\n<\/p>\n","protected":false},"author":264444,"menu_order":14,"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-18335","chapter","type-chapter","status-publish","hentry"],"part":18142,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18335","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":26,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18335\/revisions"}],"predecessor-version":[{"id":18709,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18335\/revisions\/18709"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/18142"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18335\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=18335"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=18335"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=18335"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=18335"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}