{"id":108,"date":"2023-06-21T13:22:34","date_gmt":"2023-06-21T13:22:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solving-other-types-of-equations\/"},"modified":"2023-09-07T16:31:29","modified_gmt":"2023-09-07T16:31:29","slug":"solving-other-types-of-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solving-other-types-of-equations\/","title":{"raw":"\u25aa   Solving Other Types of Equations","rendered":"\u25aa   Solving Other Types of Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve polynomial equations.<\/li>\r\n \t<li>Solve absolute value equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.\r\n<div class=\"textbox examples\">\r\n<h3>recall the zero-product property<\/h3>\r\nSolving by factoring depends on the <strong>zero-product property\u00a0<\/strong>which states that if [latex]a \\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions.\r\n\r\nIn other words if the product of two expressions equals zero, then at least one of the expressions must equal zero.\r\n\r\nWe may apply the zero-product property to polynomial equations in the same way we did to quadratic equations.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Polynomial Equations<\/h3>\r\nA polynomial of degree <em>n <\/em>is an expression of the type\r\n<div style=\"text-align: center;\">[latex]{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\\cdot \\cdot \\cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}[\/latex]<\/div>\r\nwhere <em>n<\/em> is a positive integer and [latex]{a}_{n},\\dots ,{a}_{0}[\/latex] are real numbers and [latex]{a}_{n}\\ne 0[\/latex].\r\n\r\nSetting the polynomial equal to zero gives a <strong>polynomial equation<\/strong>. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent <em>n<\/em>.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"245\"]34186[\/ohm_question]\r\n\r\n<\/div>\r\n<h2><span style=\"color: #ff9900;\">[Optional]<\/span> Solving an Absolute Value Equation<\/h2>\r\nAn <strong>absolute value equation<\/strong> is an equation in which the variable of interest is contained within absolute value bars. Absolute value is defined as a distance. That is, the bars are used to designate that the number inside the absolute value represents its distance from zero on the number line. For example, to solve an equation such as [latex]\\mid2x - 6\\mid=8,[\/latex] notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[\/latex] or [latex]-8[\/latex]. This leads to two different equations we can solve independently.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}2x - 6=8\\hfill &amp; \\text{ or }\\hfill &amp; 2x - 6=-8\\hfill \\\\ 2x=14\\hfill &amp; \\hfill &amp; 2x=-2\\hfill \\\\ x=7\\hfill &amp; \\hfill &amp; x=-1\\hfill \\end{array}[\/latex]<\/div>\r\nChecking both solutions by evaluating them in the original equation shows that both [latex]7[\/latex] and [latex]-1[\/latex] both make the statement true.\r\n\r\nIn applications, we may need to identify numbers or points on a line that are a specified distance from a given reference point. We can use an absolute value to accomplish such a task.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Absolute Value Equations<\/h3>\r\nThe absolute value of <em>x <\/em>is written as [latex]|x|[\/latex]. It has the following properties:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If } x\\ge 0,\\text{ then }|x|=x.\\hfill \\\\ \\text{If }x&lt;0,\\text{ then }|x|=-x.\\hfill \\end{array}[\/latex]<\/div>\r\nFor real numbers [latex]A[\/latex] and [latex]B[\/latex], an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex], will have solutions when [latex]A=B[\/latex] or [latex]A=-B[\/latex]. If [latex]B&lt;0[\/latex], the equation [latex]|A|=B[\/latex] has no solution.\r\n\r\nAn <strong>absolute value equation<\/strong> in the form [latex]|ax+b|=c[\/latex] has the following properties:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If }c&lt;0,|ax+b|=c\\text{ has no solution}.\\hfill \\\\ \\text{If }c=0,|ax+b|=c\\text{ has one solution}.\\hfill \\\\ \\text{If }c&gt;0,|ax+b|=c\\text{ has two solutions}.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an absolute value equation, solve it<\/h3>\r\n<ol>\r\n \t<li>Isolate the absolute value expression on one side of the equal sign.<\/li>\r\n \t<li>If [latex]c&gt;0[\/latex], write and solve two equations: [latex]ax+b=c[\/latex] and [latex]ax+b=-c[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Absolute Value Equations<\/h3>\r\nSolve the following absolute value equations:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]|6x+4|=8[\/latex]<\/li>\r\n \t<li>[latex]|3x+4|=-9[\/latex]<\/li>\r\n \t<li>[latex]|3x - 5|-4=6[\/latex]<\/li>\r\n \t<li>[latex]|-5x+10|=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"67591\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"67591\"]\r\n\r\na. [latex]|6x+4|=8[\/latex]\r\n\r\nWrite two equations and solve each:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}6x+4=8\\hfill &amp; \\text{ or } &amp; 6x+4=-8\\hfill &amp; \\\\ 6x=4\\hfill &amp; \\hfill &amp; 6x=-12\\hfill &amp; \\\\ x=\\frac{2}{3}\\hfill&amp; \\hfill &amp; x=-2\\hfill \\end{array}[\/latex]<\/p>\r\nThe two solutions are [latex]x=\\frac{2}{3}[\/latex], [latex]x=-2[\/latex].\r\n\r\nb. [latex]|3x+4|=-9[\/latex]\r\n\r\nThere is no solution as an absolute value cannot be negative.\r\n\r\nc. [latex]|3x - 5|-4=6[\/latex]\r\n\r\nIsolate the absolute value expression and then write two equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill &amp; |3x - 5|-4=6\\hfill &amp; \\hfill \\\\ \\hfill &amp; |3x - 5|=10\\hfill &amp; \\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill \\\\ 3x - 5=10\\hfill &amp; \\hfill &amp; 3x - 5=-10\\hfill \\\\ 3x=15\\hfill &amp; \\hfill &amp; 3x=-5\\hfill \\\\ x=5\\hfill &amp; \\hfill &amp; x=-\\frac{5}{3}\\hfill \\end{array}[\/latex]<\/div>\r\nThere are two solutions: [latex]x=5[\/latex], [latex]x=-\\frac{5}{3}[\/latex].\r\n\r\nd. [latex]|-5x+10|=0[\/latex]\r\n\r\nThe equation is set equal to zero, so we have to write only one equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}-5x+10=0\\hfill &amp; \\\\ -5x=-10\\hfill &amp; \\\\ x=2\\hfill \\end{array}[\/latex]<\/div>\r\nThere is one solution: [latex]x=2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>recall properties of equality<\/h3>\r\nWhen solving absolute value equations, the absolute value expression must be isolated on one side of the equation\u00a0<strong>before<\/strong> setting up the two cases to remove the absolute value bars.\r\n\r\nUse the properties of equality to isolate the absolute value expression, but avoid multiplying into or dividing from any expression inside the bars.\r\n\r\nEx. use the properties of equality and the techniques of solving linear equations to isolate the absolute value expression in the equation [latex]5-2\\mid3x-4\\mid=-7[\/latex], then solve the equation.\r\n\r\n[reveal-answer q=\"623732\"]more[\/reveal-answer]\r\n[hidden-answer a=\"623732\"]\r\n\r\n[latex]\\begin{align}5-2\\mid3x-4\\mid &amp;= -7 \\\\ -2\\mid3x-4\\mid &amp;= -12 \\quad \\text{subtract the 5 to the right-hand side} \\\\ \\mid3x-4\\mid &amp;= 6 \\qquad \\ \\text{divide by -2} \\end{align} [\/latex]\r\n\r\n&nbsp;\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\"><span style=\"text-decoration: underline;\">Solution<\/span>\u00a0<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[latex] x=- \\dfrac{2}{3}, x= \\dfrac{10}{3}[\/latex]<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the absolute value equation: [latex]|1 - 4x|+8=13[\/latex].\r\n\r\n[reveal-answer q=\"567620\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567620\"]\r\n\r\n[latex]x=-1[\/latex], [latex]x=\\frac{3}{2}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question height=\"215\"]60839[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Other Types of Equations<\/h2>\r\nThere are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.\r\n<h2>Solving Equations in Quadratic Form<\/h2>\r\n<strong>Equations in quadratic form <\/strong>are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[\/latex], and [latex]{x}^{\\frac{2}{3}}+4{x}^{\\frac{1}{3}}+2=0[\/latex]. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Quadratic Form<\/h3>\r\nIf the exponent on the middle term is one-half of the exponent on the leading term, we have an <strong>equation in quadratic form\u00a0<\/strong>which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation quadratic in form, solve it<\/h3>\r\n<ol>\r\n \t<li>Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.<\/li>\r\n \t<li>If it is, substitute a variable, such as <em>u<\/em>, for the variable portion of the middle term.<\/li>\r\n \t<li>Rewrite the equation so that it takes on the standard form of a quadratic.<\/li>\r\n \t<li>Solve using one of the usual methods for solving a quadratic.<\/li>\r\n \t<li>Replace the substitution variable with the original term.<\/li>\r\n \t<li>Solve the remaining equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Fourth-Degree Equation in Quadratic Form<\/h3>\r\nSolve this fourth-degree equation: [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex].\r\n\r\n[reveal-answer q=\"82705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"82705\"]\r\n\r\nThis equation fits the main criteria: that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let [latex]u={x}^{2}[\/latex]. Rewrite the equation in [latex]u[\/latex].\r\n<div style=\"text-align: center;\">[latex]3{u}^{2}-13u+4=0[\/latex]<\/div>\r\nNow solve the quadratic.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{u}^{2}-13u + 4=0\\hfill &amp; \\\\ \\left(3u-1\\right)\\left(u - 4\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nSolve for [latex]u[\/latex]<em>.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}3u-1=0\\hfill &amp; \\\\ 3u=1\\hfill &amp; \\\\ u=\\frac{1}{3}\\hfill &amp; \\\\ {x}^{2}=\\frac{1}{3}\\hfill &amp; \\\\ x=\\pm \\sqrt{\\frac{1}{3}}\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: left;\">Replace\u00a0[latex]u[\/latex]\u00a0with its original term.<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 4=0\\hfill &amp; \\\\ u=4\\hfill &amp; \\\\ {x}^{2}=4\\hfill &amp; \\\\ x=\\pm 2\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=\\pm \\sqrt{\\frac{1}{3}}[\/latex] and [latex]x=\\pm {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 using substitution: [latex]{x}^{4}-6{x}^{2}+5=0[\/latex].\r\n\r\n[reveal-answer q=\"934781\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"934781\"]\r\n\r\n[latex]x=1,-1,\\sqrt{5},-\\sqrt{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation in Quadratic Form Containing a Binomial<\/h3>\r\nSolve the equation in quadratic form: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex].\r\n\r\n[reveal-answer q=\"330904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"330904\"]\r\n\r\nThis equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution letting [latex]u=x+2[\/latex]. Then rewrite the equation in <em>u.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{u}^{2}+11u - 12=0\\hfill &amp; \\\\ \\left(u+12\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nSolve using the zero-factor property and then replace <em>u<\/em> with the original expression.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u+12=0\\hfill &amp; \\\\ u=-12\\hfill &amp; \\\\ x+2=-12\\hfill &amp; \\\\ x=-14\\hfill \\end{array}[\/latex]<\/div>\r\nThe second factor results in\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill &amp; \\\\ u=1\\hfill &amp; \\\\ x+2=1\\hfill &amp; \\\\ x=-1\\hfill \\end{array}[\/latex]<\/div>\r\nWe have two solutions: [latex]x=-14[\/latex], [latex]x=-1[\/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: [latex]{\\left(x - 5\\right)}^{2}-4\\left(x - 5\\right)-21=0[\/latex].\r\n\r\n[reveal-answer q=\"737690\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737690\"]\r\n\r\n[latex]x=2,x=12[\/latex][\/hidden-answer]\r\n\r\n[ohm_question height=\"215\"]1883[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Rational Equations Resulting in a Quadratic<\/h2>\r\nEarlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.\r\n<div class=\"textbox examples\">\r\n<h3>Recall division by zero<\/h3>\r\nDividing zero by a number results in zero.\r\n<p style=\"padding-left: 30px;\">[latex]\\dfrac{0}{a} = 0[\/latex].<\/p>\r\nBut we cannot divide by zero. A zero in a denominator results in an\u00a0<em>undefined<\/em> expression.\r\n<p style=\"padding-left: 30px;\">[latex]\\dfrac{b}{0}=\\text{ undefined}[\/latex]<\/p>\r\nWhen solving rational equations in which a variable is contained in a denominator, always declare the value of the variable which would result in an undefined equation so that you won't forget to exclude it from the solution set.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\r\nSolve the following rational equation: [latex]\\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].\r\n\r\n[reveal-answer q=\"164755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164755\"]\r\n\r\nWe want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right) \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)=-8 \\\\ -4{x}^{2}-4x+4x - 4=-8 \\\\ -4{x}^{2}+4=0 \\\\ -4\\left({x}^{2}-1\\right)=0 \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)=0 \\\\ x=-1 \\\\ x=1 \\end{array}[\/latex]<\/div>\r\nIn this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.\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 [latex]\\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].\r\n\r\n[reveal-answer q=\"586054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"586054\"]\r\n\r\n[latex]x=-1[\/latex], [latex]x=0[\/latex] is not a solution.[\/hidden-answer]\r\n\r\n[ohm_question height=\"200\"]3496[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve polynomial equations.<\/li>\n<li>Solve absolute value equations.<\/li>\n<\/ul>\n<\/div>\n<p>We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.<\/p>\n<div class=\"textbox examples\">\n<h3>recall the zero-product property<\/h3>\n<p>Solving by factoring depends on the <strong>zero-product property\u00a0<\/strong>which states that if [latex]a \\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions.<\/p>\n<p>In other words if the product of two expressions equals zero, then at least one of the expressions must equal zero.<\/p>\n<p>We may apply the zero-product property to polynomial equations in the same way we did to quadratic equations.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Polynomial Equations<\/h3>\n<p>A polynomial of degree <em>n <\/em>is an expression of the type<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\\cdot \\cdot \\cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}[\/latex]<\/div>\n<p>where <em>n<\/em> is a positive integer and [latex]{a}_{n},\\dots ,{a}_{0}[\/latex] are real numbers and [latex]{a}_{n}\\ne 0[\/latex].<\/p>\n<p>Setting the polynomial equal to zero gives a <strong>polynomial equation<\/strong>. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent <em>n<\/em>.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm34186\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=34186&theme=oea&iframe_resize_id=ohm34186&show_question_numbers\" width=\"100%\" height=\"245\"><\/iframe><\/p>\n<\/div>\n<h2><span style=\"color: #ff9900;\">[Optional]<\/span> Solving an Absolute Value Equation<\/h2>\n<p>An <strong>absolute value equation<\/strong> is an equation in which the variable of interest is contained within absolute value bars. Absolute value is defined as a distance. That is, the bars are used to designate that the number inside the absolute value represents its distance from zero on the number line. For example, to solve an equation such as [latex]\\mid2x - 6\\mid=8,[\/latex] notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[\/latex] or [latex]-8[\/latex]. This leads to two different equations we can solve independently.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}2x - 6=8\\hfill & \\text{ or }\\hfill & 2x - 6=-8\\hfill \\\\ 2x=14\\hfill & \\hfill & 2x=-2\\hfill \\\\ x=7\\hfill & \\hfill & x=-1\\hfill \\end{array}[\/latex]<\/div>\n<p>Checking both solutions by evaluating them in the original equation shows that both [latex]7[\/latex] and [latex]-1[\/latex] both make the statement true.<\/p>\n<p>In applications, we may need to identify numbers or points on a line that are a specified distance from a given reference point. We can use an absolute value to accomplish such a task.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value Equations<\/h3>\n<p>The absolute value of <em>x <\/em>is written as [latex]|x|[\/latex]. It has the following properties:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If } x\\ge 0,\\text{ then }|x|=x.\\hfill \\\\ \\text{If }x<0,\\text{ then }|x|=-x.\\hfill \\end{array}[\/latex]<\/div>\n<p>For real numbers [latex]A[\/latex] and [latex]B[\/latex], an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex], will have solutions when [latex]A=B[\/latex] or [latex]A=-B[\/latex]. If [latex]B<0[\/latex], the equation [latex]|A|=B[\/latex] has no solution.\n\nAn <strong>absolute value equation<\/strong> in the form [latex]|ax+b|=c[\/latex] has the following properties:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If }c<0,|ax+b|=c\\text{ has no solution}.\\hfill \\\\ \\text{If }c=0,|ax+b|=c\\text{ has one solution}.\\hfill \\\\ \\text{If }c>0,|ax+b|=c\\text{ has two solutions}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an absolute value equation, solve it<\/h3>\n<ol>\n<li>Isolate the absolute value expression on one side of the equal sign.<\/li>\n<li>If [latex]c>0[\/latex], write and solve two equations: [latex]ax+b=c[\/latex] and [latex]ax+b=-c[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Absolute Value Equations<\/h3>\n<p>Solve the following absolute value equations:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]|6x+4|=8[\/latex]<\/li>\n<li>[latex]|3x+4|=-9[\/latex]<\/li>\n<li>[latex]|3x - 5|-4=6[\/latex]<\/li>\n<li>[latex]|-5x+10|=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q67591\">Show Solution<\/span><\/p>\n<div id=\"q67591\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]|6x+4|=8[\/latex]<\/p>\n<p>Write two equations and solve each:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}6x+4=8\\hfill & \\text{ or } & 6x+4=-8\\hfill & \\\\ 6x=4\\hfill & \\hfill & 6x=-12\\hfill & \\\\ x=\\frac{2}{3}\\hfill& \\hfill & x=-2\\hfill \\end{array}[\/latex]<\/p>\n<p>The two solutions are [latex]x=\\frac{2}{3}[\/latex], [latex]x=-2[\/latex].<\/p>\n<p>b. [latex]|3x+4|=-9[\/latex]<\/p>\n<p>There is no solution as an absolute value cannot be negative.<\/p>\n<p>c. [latex]|3x - 5|-4=6[\/latex]<\/p>\n<p>Isolate the absolute value expression and then write two equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill & |3x - 5|-4=6\\hfill & \\hfill \\\\ \\hfill & |3x - 5|=10\\hfill & \\hfill \\\\ \\hfill & \\hfill & \\hfill \\\\ 3x - 5=10\\hfill & \\hfill & 3x - 5=-10\\hfill \\\\ 3x=15\\hfill & \\hfill & 3x=-5\\hfill \\\\ x=5\\hfill & \\hfill & x=-\\frac{5}{3}\\hfill \\end{array}[\/latex]<\/div>\n<p>There are two solutions: [latex]x=5[\/latex], [latex]x=-\\frac{5}{3}[\/latex].<\/p>\n<p>d. [latex]|-5x+10|=0[\/latex]<\/p>\n<p>The equation is set equal to zero, so we have to write only one equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}-5x+10=0\\hfill & \\\\ -5x=-10\\hfill & \\\\ x=2\\hfill \\end{array}[\/latex]<\/div>\n<p>There is one solution: [latex]x=2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall properties of equality<\/h3>\n<p>When solving absolute value equations, the absolute value expression must be isolated on one side of the equation\u00a0<strong>before<\/strong> setting up the two cases to remove the absolute value bars.<\/p>\n<p>Use the properties of equality to isolate the absolute value expression, but avoid multiplying into or dividing from any expression inside the bars.<\/p>\n<p>Ex. use the properties of equality and the techniques of solving linear equations to isolate the absolute value expression in the equation [latex]5-2\\mid3x-4\\mid=-7[\/latex], then solve the equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q623732\">more<\/span><\/p>\n<div id=\"q623732\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}5-2\\mid3x-4\\mid &= -7 \\\\ -2\\mid3x-4\\mid &= -12 \\quad \\text{subtract the 5 to the right-hand side} \\\\ \\mid3x-4\\mid &= 6 \\qquad \\ \\text{divide by -2} \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><span style=\"text-decoration: underline;\">Solution<\/span>\u00a0<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">[latex]x=- \\dfrac{2}{3}, x= \\dfrac{10}{3}[\/latex]<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the absolute value equation: [latex]|1 - 4x|+8=13[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567620\">Show Solution<\/span><\/p>\n<div id=\"q567620\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1[\/latex], [latex]x=\\frac{3}{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm60839\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=60839&theme=oea&iframe_resize_id=ohm60839&show_question_numbers\" width=\"100%\" height=\"215\"><\/iframe><\/p>\n<\/div>\n<h2>Other Types of Equations<\/h2>\n<p>There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.<\/p>\n<h2>Solving Equations in Quadratic Form<\/h2>\n<p><strong>Equations in quadratic form <\/strong>are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[\/latex], and [latex]{x}^{\\frac{2}{3}}+4{x}^{\\frac{1}{3}}+2=0[\/latex]. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Quadratic Form<\/h3>\n<p>If the exponent on the middle term is one-half of the exponent on the leading term, we have an <strong>equation in quadratic form\u00a0<\/strong>which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation quadratic in form, solve it<\/h3>\n<ol>\n<li>Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.<\/li>\n<li>If it is, substitute a variable, such as <em>u<\/em>, for the variable portion of the middle term.<\/li>\n<li>Rewrite the equation so that it takes on the standard form of a quadratic.<\/li>\n<li>Solve using one of the usual methods for solving a quadratic.<\/li>\n<li>Replace the substitution variable with the original term.<\/li>\n<li>Solve the remaining equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Fourth-Degree Equation in Quadratic Form<\/h3>\n<p>Solve this fourth-degree equation: [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q82705\">Show Solution<\/span><\/p>\n<div id=\"q82705\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation fits the main criteria: that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let [latex]u={x}^{2}[\/latex]. Rewrite the equation in [latex]u[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]3{u}^{2}-13u+4=0[\/latex]<\/div>\n<p>Now solve the quadratic.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{u}^{2}-13u + 4=0\\hfill & \\\\ \\left(3u-1\\right)\\left(u - 4\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Solve for [latex]u[\/latex]<em>.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}3u-1=0\\hfill & \\\\ 3u=1\\hfill & \\\\ u=\\frac{1}{3}\\hfill & \\\\ {x}^{2}=\\frac{1}{3}\\hfill & \\\\ x=\\pm \\sqrt{\\frac{1}{3}}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div style=\"text-align: left;\">Replace\u00a0[latex]u[\/latex]\u00a0with its original term.<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 4=0\\hfill & \\\\ u=4\\hfill & \\\\ {x}^{2}=4\\hfill & \\\\ x=\\pm 2\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=\\pm \\sqrt{\\frac{1}{3}}[\/latex] and [latex]x=\\pm {2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve using substitution: [latex]{x}^{4}-6{x}^{2}+5=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934781\">Show Answer<\/span><\/p>\n<div id=\"q934781\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=1,-1,\\sqrt{5},-\\sqrt{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation in Quadratic Form Containing a Binomial<\/h3>\n<p>Solve the equation in quadratic form: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q330904\">Show Solution<\/span><\/p>\n<div id=\"q330904\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution letting [latex]u=x+2[\/latex]. Then rewrite the equation in <em>u.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{u}^{2}+11u - 12=0\\hfill & \\\\ \\left(u+12\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Solve using the zero-factor property and then replace <em>u<\/em> with the original expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u+12=0\\hfill & \\\\ u=-12\\hfill & \\\\ x+2=-12\\hfill & \\\\ x=-14\\hfill \\end{array}[\/latex]<\/div>\n<p>The second factor results in<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill & \\\\ u=1\\hfill & \\\\ x+2=1\\hfill & \\\\ x=-1\\hfill \\end{array}[\/latex]<\/div>\n<p>We have two solutions: [latex]x=-14[\/latex], [latex]x=-1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve: [latex]{\\left(x - 5\\right)}^{2}-4\\left(x - 5\\right)-21=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737690\">Show Solution<\/span><\/p>\n<div id=\"q737690\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=2,x=12[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1883\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1883&theme=oea&iframe_resize_id=ohm1883&show_question_numbers\" width=\"100%\" height=\"215\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Rational Equations Resulting in a Quadratic<\/h2>\n<p>Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall division by zero<\/h3>\n<p>Dividing zero by a number results in zero.<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\dfrac{0}{a} = 0[\/latex].<\/p>\n<p>But we cannot divide by zero. A zero in a denominator results in an\u00a0<em>undefined<\/em> expression.<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\dfrac{b}{0}=\\text{ undefined}[\/latex]<\/p>\n<p>When solving rational equations in which a variable is contained in a denominator, always declare the value of the variable which would result in an undefined equation so that you won&#8217;t forget to exclude it from the solution set.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\n<p>Solve the following rational equation: [latex]\\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164755\">Show Solution<\/span><\/p>\n<div id=\"q164755\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right) \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)=-8 \\\\ -4{x}^{2}-4x+4x - 4=-8 \\\\ -4{x}^{2}+4=0 \\\\ -4\\left({x}^{2}-1\\right)=0 \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)=0 \\\\ x=-1 \\\\ x=1 \\end{array}[\/latex]<\/div>\n<p>In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q586054\">Show Solution<\/span><\/p>\n<div id=\"q586054\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1[\/latex], [latex]x=0[\/latex] is not a solution.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm3496\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3496&theme=oea&iframe_resize_id=ohm3496&show_question_numbers\" width=\"100%\" height=\"200\"><\/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-108\">\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 34186. <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 60839. <strong>Authored by<\/strong>: Alyson Day. <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 1883. <strong>Authored by<\/strong>: Barbara Goldner. <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 3496. <strong>Authored by<\/strong>: Shawn Triplett. <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 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