{"id":2613,"date":"2016-11-04T17:56:38","date_gmt":"2016-11-04T17:56:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2613"},"modified":"2018-07-28T00:35:29","modified_gmt":"2018-07-28T00:35:29","slug":"introduction-quadratic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/chapter\/introduction-quadratic-equations\/","title":{"raw":"Quadratic Equations","rendered":"Quadratic Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\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>Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle.<\/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 \t<li>Use the discriminant to determine the number and type of solutions to a quadratic equation.<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe left computer monitor in the image below is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200357\/CNX_CAT_Figure_02_05_001.jpg\" alt=\"Two televisions side-by-side. The right television is slightly larger than the left. \" width=\"731\" height=\"304\" \/>\r\n<h2>Factoring and the Square Root Property<\/h2>\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\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<h3>Solving Quadratics with a Leading Coefficient of 1<\/h3>\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<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2029&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using the Square Root Property<\/h3>\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\">\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][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29172&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using the Pythagorean Theorem<\/h3>\r\nOne of the most famous formulas in mathematics is the <strong>Pythagorean Theorem<\/strong>. It is based on a right triangle and states the relationship among the lengths of the sides as [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] refer to the legs of a right triangle adjacent to the [latex]90^\\circ [\/latex] angle, and [latex]c[\/latex] refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.\r\n\r\nWe use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.\r\n\r\nThe Pythagorean Theorem is given as\r\n<div>[latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex]<\/div>\r\nwhere [latex]a[\/latex] and [latex]b[\/latex] refer to the legs of a right triangle adjacent to the [latex]{90}^{\\circ }[\/latex] angle, and [latex]c[\/latex] refers to the hypotenuse.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24224532\/CNX_CAT_Figure_02_05_004.jpg\" alt=\"Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c\" width=\"487\" height=\"196\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Length of the Missing Side of a Right Triangle<\/h3>\r\nFind the length of the missing side of the right triangle.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24224534\/CNX_CAT_Figure_02_05_005.jpg\" alt=\"Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.\" width=\"487\" height=\"112\" \/>\r\n[reveal-answer q=\"111347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111347\"]\r\n\r\nAs we have measurements for side <em>b<\/em> and the hypotenuse, the missing side is <em>a.<\/em>\r\n<div>[latex]\\begin{array}{l}{a}^{2}+{b}^{2}={c}^{2}\\hfill \\\\ {a}^{2}+{\\left(4\\right)}^{2}={\\left(12\\right)}^{2}\\hfill \\\\ {a}^{2}+16=144\\hfill \\\\ {a}^{2}=128\\hfill \\\\ a=\\sqrt{128}\\hfill \\\\ a=8\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the Pythagorean Theorem to solve the right triangle problem: Leg <em>a <\/em>measures 4 units, leg <em>b <\/em>measures 3 units. Find the length of the hypotenuse.\r\n\r\n[reveal-answer q=\"863223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"863223\"]\r\n\r\n[latex]5[\/latex] units[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=48710&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Completing the Square and the Quadratic Formula<\/h2>\r\nNot all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n\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 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<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1384&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79619&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Using the Quadratic Formula<\/h3>\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 the 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 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;=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&amp;=\\frac{-\\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;=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&amp;=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&amp;=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].[\/hidden-answer]. Notice they are written in standard form of a complex number. When a solution is a complex number, you must separate the real part from the imaginary part and write it in standard form.\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<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4014&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35639&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>The Discriminant<\/h3>\r\nThe <strong>quadratic formula<\/strong> not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the <strong>discriminant<\/strong>, or the expression under the radical, [latex]{b}^{2}-4ac[\/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers as well as how many solutions of each type to expect. The table below\u00a0relates the value of the discriminant to the solutions of a quadratic equation.\r\n<table summary=\"A table with 5 rows and 2 columns. The entries in the first row are: Value of Discriminant and Results. The entries in the second row are: b squared minus four times a times c equals zero and One rational solution (double solution). The entries in the third row are: b squared minus four times a times c is greater than zero, perfect square and Two rational solutions. The entries in the fourth row are: b squared minus four times a times c is greater than zero, not a perfect square and Two irrational solutions. The entries in the fifth row are: b squared minus four times a times c is less than zero and Two complex solutions.\">\r\n<thead>\r\n<tr>\r\n<th>Value of Discriminant<\/th>\r\n<th>Results<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac=0[\/latex]<\/td>\r\n<td>One rational solution (double solution)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&gt;0[\/latex], perfect square<\/td>\r\n<td>Two rational solutions<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&gt;0[\/latex], not a perfect square<\/td>\r\n<td>Two irrational solutions<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&lt;0[\/latex]<\/td>\r\n<td>Two complex solutions<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Discriminant<\/h3>\r\nFor [latex]a{x}^{2}+bx+c=0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers, the <strong>discriminant<\/strong> is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[\/latex]. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation<\/h3>\r\nUse the discriminant to find the nature of the solutions to the following quadratic equations:\r\n<ol>\r\n \t<li>[latex]{x}^{2}+4x+4=0[\/latex]<\/li>\r\n \t<li>[latex]8{x}^{2}+14x+3=0[\/latex]<\/li>\r\n \t<li>[latex]3{x}^{2}-5x - 2=0[\/latex]<\/li>\r\n \t<li>[latex]3{x}^{2}-10x+15=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"229118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229118\"]\r\n\r\nCalculate the discriminant [latex]{b}^{2}-4ac[\/latex] for each equation and state the expected type of solutions.\r\n<ol>\r\n \t<li>[latex]{x}^{2}+4x+4=0[\/latex]: [latex]{b}^{2}-4ac={\\left(4\\right)}^{2}-4\\left(1\\right)\\left(4\\right)=0[\/latex]. There will be one rational double solution.<\/li>\r\n \t<li>[latex]8{x}^{2}+14x+3=0[\/latex]: [latex]{b}^{2}-4ac={\\left(14\\right)}^{2}-4\\left(8\\right)\\left(3\\right)=100[\/latex]. As [latex]100[\/latex] is a perfect square, there will be two rational solutions.<\/li>\r\n \t<li>[latex]3{x}^{2}-5x - 2=0[\/latex]: [latex]{b}^{2}-4ac={\\left(-5\\right)}^{2}-4\\left(3\\right)\\left(-2\\right)=49[\/latex]. As [latex]49[\/latex] is a perfect square, there will be two rational solutions.<\/li>\r\n \t<li>[latex]3{x}^{2}-10x+15=0[\/latex]: [latex]{b}^{2}-4ac={\\left(-10\\right)}^{2}-4\\left(3\\right)\\left(15\\right)=-80[\/latex]. There will be two complex solutions.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35145&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"350\"><\/iframe>\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>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\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>Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle.<\/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<li>Use the discriminant to determine the number and type of solutions to a quadratic equation.<\/li>\n<\/ul>\n<\/div>\n<p>The left computer monitor in the image below is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200357\/CNX_CAT_Figure_02_05_001.jpg\" alt=\"Two televisions side-by-side. The right television is slightly larger than the left.\" width=\"731\" height=\"304\" \/><\/p>\n<h2>Factoring and the Square Root Property<\/h2>\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<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<h3>Solving Quadratics with a Leading Coefficient of 1<\/h3>\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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2029&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Square Root Property<\/h3>\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\">\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><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29172&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Pythagorean Theorem<\/h3>\n<p>One of the most famous formulas in mathematics is the <strong>Pythagorean Theorem<\/strong>. It is based on a right triangle and states the relationship among the lengths of the sides as [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] refer to the legs of a right triangle adjacent to the [latex]90^\\circ[\/latex] angle, and [latex]c[\/latex] refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.<\/p>\n<p>We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.<\/p>\n<p>The Pythagorean Theorem is given as<\/p>\n<div>[latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex]<\/div>\n<p>where [latex]a[\/latex] and [latex]b[\/latex] refer to the legs of a right triangle adjacent to the [latex]{90}^{\\circ }[\/latex] angle, and [latex]c[\/latex] refers to the hypotenuse.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24224532\/CNX_CAT_Figure_02_05_004.jpg\" alt=\"Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c\" width=\"487\" height=\"196\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Length of the Missing Side of a Right Triangle<\/h3>\n<p>Find the length of the missing side of the right triangle.<\/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\/24224534\/CNX_CAT_Figure_02_05_005.jpg\" alt=\"Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.\" width=\"487\" height=\"112\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q111347\">Show Solution<\/span><\/p>\n<div id=\"q111347\" class=\"hidden-answer\" style=\"display: none\">\n<p>As we have measurements for side <em>b<\/em> and the hypotenuse, the missing side is <em>a.<\/em><\/p>\n<div>[latex]\\begin{array}{l}{a}^{2}+{b}^{2}={c}^{2}\\hfill \\\\ {a}^{2}+{\\left(4\\right)}^{2}={\\left(12\\right)}^{2}\\hfill \\\\ {a}^{2}+16=144\\hfill \\\\ {a}^{2}=128\\hfill \\\\ a=\\sqrt{128}\\hfill \\\\ a=8\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the Pythagorean Theorem to solve the right triangle problem: Leg <em>a <\/em>measures 4 units, leg <em>b <\/em>measures 3 units. Find the length of the hypotenuse.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q863223\">Show Solution<\/span><\/p>\n<div id=\"q863223\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]5[\/latex] units<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=48710&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h2>Completing the Square and the Quadratic Formula<\/h2>\n<p>Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\n<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 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><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1384&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79619&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Quadratic Formula<\/h3>\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 the 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 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&=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&=\\frac{-\\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&=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].<\/p><\/div>\n<\/div>\n<p>. Notice they are written in standard form of a complex number. When a solution is a complex number, you must separate the real part from the imaginary part and write it in standard form.<\/p>\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=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4014&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35639&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>The Discriminant<\/h3>\n<p>The <strong>quadratic formula<\/strong> not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the <strong>discriminant<\/strong>, or the expression under the radical, [latex]{b}^{2}-4ac[\/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers as well as how many solutions of each type to expect. The table below\u00a0relates the value of the discriminant to the solutions of a quadratic equation.<\/p>\n<table summary=\"A table with 5 rows and 2 columns. The entries in the first row are: Value of Discriminant and Results. The entries in the second row are: b squared minus four times a times c equals zero and One rational solution (double solution). The entries in the third row are: b squared minus four times a times c is greater than zero, perfect square and Two rational solutions. The entries in the fourth row are: b squared minus four times a times c is greater than zero, not a perfect square and Two irrational solutions. The entries in the fifth row are: b squared minus four times a times c is less than zero and Two complex solutions.\">\n<thead>\n<tr>\n<th>Value of Discriminant<\/th>\n<th>Results<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]{b}^{2}-4ac=0[\/latex]<\/td>\n<td>One rational solution (double solution)<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac>0[\/latex], perfect square<\/td>\n<td>Two rational solutions<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac>0[\/latex], not a perfect square<\/td>\n<td>Two irrational solutions<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac<0[\/latex]<\/td>\n<td>Two complex solutions<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox\">\n<h3>A General Note: The Discriminant<\/h3>\n<p>For [latex]a{x}^{2}+bx+c=0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers, the <strong>discriminant<\/strong> is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[\/latex]. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation<\/h3>\n<p>Use the discriminant to find the nature of the solutions to the following quadratic equations:<\/p>\n<ol>\n<li>[latex]{x}^{2}+4x+4=0[\/latex]<\/li>\n<li>[latex]8{x}^{2}+14x+3=0[\/latex]<\/li>\n<li>[latex]3{x}^{2}-5x - 2=0[\/latex]<\/li>\n<li>[latex]3{x}^{2}-10x+15=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229118\">Show Solution<\/span><\/p>\n<div id=\"q229118\" class=\"hidden-answer\" style=\"display: none\">\n<p>Calculate the discriminant [latex]{b}^{2}-4ac[\/latex] for each equation and state the expected type of solutions.<\/p>\n<ol>\n<li>[latex]{x}^{2}+4x+4=0[\/latex]: [latex]{b}^{2}-4ac={\\left(4\\right)}^{2}-4\\left(1\\right)\\left(4\\right)=0[\/latex]. There will be one rational double solution.<\/li>\n<li>[latex]8{x}^{2}+14x+3=0[\/latex]: [latex]{b}^{2}-4ac={\\left(14\\right)}^{2}-4\\left(8\\right)\\left(3\\right)=100[\/latex]. As [latex]100[\/latex] is a perfect square, there will be two rational solutions.<\/li>\n<li>[latex]3{x}^{2}-5x - 2=0[\/latex]: [latex]{b}^{2}-4ac={\\left(-5\\right)}^{2}-4\\left(3\\right)\\left(-2\\right)=49[\/latex]. As [latex]49[\/latex] is a perfect square, there will be two rational solutions.<\/li>\n<li>[latex]3{x}^{2}-10x+15=0[\/latex]: [latex]{b}^{2}-4ac={\\left(-10\\right)}^{2}-4\\left(3\\right)\\left(15\\right)=-80[\/latex]. There will be two complex solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35145&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"350\"><\/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\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-2613\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><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@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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@5.2<\/li><li>Question ID 2029. <strong>Authored by<\/strong>: Lawrence Morales. <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>: IMathAS Community License CC-BY + GPL<\/li><li>Question 29172, 35145. <strong>Authored by<\/strong>: Jim Smart. <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>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 48710. <strong>Authored by<\/strong>: Darlene Diaz. <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>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 1384. <strong>Authored by<\/strong>: WebWork-Rochester. <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>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 79619. <strong>Authored by<\/strong>: Edward Wicks. <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>: IMathAS Community License CC-BY + GPL<\/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":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 2029\",\"author\":\"Lawrence Morales\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question 29172, 35145\",\"author\":\"Jim 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