{"id":103,"date":"2023-06-21T13:22:33","date_gmt":"2023-06-21T13:22:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/factoring-and-the-square-root-property\/"},"modified":"2023-09-07T16:04:42","modified_gmt":"2023-09-07T16:04:42","slug":"factoring-and-the-square-root-property","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/factoring-and-the-square-root-property\/","title":{"raw":"\u25aa   Factoring and the Square Root Property","rendered":"\u25aa   Factoring and the Square Root Property"},"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>Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle.<\/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 height=\"220\"]2029[\/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 height=\"395\"]29172[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using the Pythagorean Theorem<\/h2>\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\r\n[ohm_question height=\"405\"]48710[\/ohm_question]\r\n\r\n<\/div>","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>Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle.<\/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=\"220\"><\/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=\"395\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Pythagorean Theorem<\/h2>\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=\"ohm48710\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=48710&theme=oea&iframe_resize_id=ohm48710&show_question_numbers\" width=\"100%\" height=\"405\"><\/iframe><\/p>\n<\/div>\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-103\">\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. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and 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