{"id":481,"date":"2024-05-01T19:14:34","date_gmt":"2024-05-01T19:14:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/?post_type=chapter&#038;p=481"},"modified":"2025-01-07T00:58:51","modified_gmt":"2025-01-07T00:58:51","slug":"6-5-solving-quadratic-inequalities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-5-solving-quadratic-inequalities\/","title":{"raw":"6.5 Solving Quadratic Inequalities","rendered":"6.5 Solving Quadratic Inequalities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve quadratic inequalities, expressing the solution set using interval notation.<\/li>\r\n<\/ul>\r\n<\/div>\r\nA <strong>quadratic inequality<\/strong> is a quadratic equation, but with the equal sign replaced by an inequality sign. An example of a quadratic inequality is\r\n<p style=\"text-align: center;\">[latex]x^2-3x-4\\leq 0[\/latex]<\/p>\r\nTo help solve this, we will consider the quadratic function [latex]f(x)=x^2-3x-4[\/latex]. Solving our inequality above is equivalent to asking which [latex]x[\/latex] values cause the output value of the function [latex]f[\/latex] to be less than or equal to [latex]0[\/latex]. We will use a graph of the function [latex]f[\/latex] to understand this idea. Below is the graph of the function [latex]f[\/latex]. Note that the function factors as [latex]f(x)=(x-4)(x+1)[\/latex], which reveals the two [latex]x[\/latex]-intercepts of [latex](-1,0)[\/latex] and [latex](4, 0)[\/latex].\r\n\r\n<img class=\"aligncenter wp-image-1878 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-271x300.png\" alt=\"A parabola is facing upwards and has x-intercepts at (-1, 0) and (4, 0)\" width=\"271\" height=\"300\" \/>\r\n\r\nTry to use the above graph to solve the inequality [latex]f(x) \\leq 0[\/latex] before revealing the answer below.\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the inequality\u00a0[latex]x^2-3x-4\\leq 0[\/latex] by graphing.\r\n\r\n[reveal-answer q=\"210097\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"210097\"]\r\n\r\nThe solution to the inequality is the portion of the graph of\u00a0[latex]f(x)=x^2-3x-4[\/latex] which has outputs less than or equal to [latex]0[\/latex], which in the graph is the portion of the curve touching or below the [latex]x[\/latex]-axis. We can determine visually that [latex]x[\/latex] values between [latex]-1[\/latex] and [latex]4[\/latex] (and including these two numbers) satisfy this condition. Here is the same graph, but with the portion below the [latex]x[\/latex]-axis colored in green and the portion above the graph dotted:\r\n\r\n<img class=\"aligncenter wp-image-1879 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-272x300.png\" alt=\"The parabola from the previous graph is marked with the portion below the x-axis highlighted in green\" width=\"272\" height=\"300\" \/>\r\n\r\nThe solution is the interval of [latex]x[\/latex]-values corresponding to the region below or touching the [latex]x[\/latex]-axis, which is the interval [latex][-1,4][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can see from the above example that the key was to use the places where the graph <em>crosses<\/em> the [latex]x[\/latex]-axis as the endpoints of our solution interval.\r\n\r\nHere is another example to try. Note the small differences in the setup of the problem.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve the inequality\u00a0[latex]x^2-6x+8 &gt; 0[\/latex].\r\n\r\n[reveal-answer q=\"908945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"908945\"]\r\n\r\nThe function [latex]f(x)=x^2-6x+8[\/latex] factors as [latex]f(x)=(x-4)(x-2)[\/latex] and so the [latex]x[\/latex]-intercepts are at [latex](4,0)[\/latex] and [latex](2,0)[\/latex]. Since the leading coefficient is positive, we draw an upward facing parabola. This gives us enough information to sketch a graph.\r\n\r\n<img class=\"aligncenter wp-image-1880 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-300x300.png\" alt=\"An upward facing parabola has x-intercepts at (2,0) and (4,0).\" width=\"300\" height=\"300\" \/>\r\n\r\nSince we are solving [latex]f(x) &gt; 0[\/latex], we are interested in the portion of the graph that is above the [latex]x[\/latex]-axis this time. Here is the same graph, but with the portion above the graph shaded green and the portion below dotted.\r\n\r\n<img class=\"aligncenter wp-image-1881 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-300x294.png\" alt=\"The previous parabola is marked with the two portions above the x-axis highlighted in green.\" width=\"300\" height=\"294\" \/>\r\n\r\nThere are two intervals of [latex]x[\/latex]-values that correspond to portions of the graph above the [latex]x[\/latex]-axis. These intervals consist of [latex]x[\/latex] values less than [latex]2[\/latex] and [latex]x[\/latex] values greater than [latex]4[\/latex], since these are the [latex]x[\/latex]-intercepts of the graph. Since we have strict inequality in this example, the values [latex]x=2[\/latex] and [latex]x=4[\/latex] are not solutions.\r\n\r\nThe solution is [latex](-\\infty, 2) \\cup (4, \\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using a Number Line<\/h2>\r\nWe can see from the above examples that the [latex]x[\/latex]-intercepts of the function associated with the inequality are crucial to the solving process and end up being the endpoints of the solution intervals. This is because a parabola is a smooth curve and can only pass from above to below the [latex]x[\/latex]-axis by passing through the axis first (resulting in an intercept).\r\n\r\nThis suggests an alternative approach to solving which does not require graphing the parabola. We can determine the \"boundary values\" where the inequality can pass from being true to false, then use test points to determine which intervals on the number line are true or false. This is similar to the test point approach used in solving systems of inequalities in Chapter 2.\r\n\r\nConsider the inequality [latex]x^2+2x-15&gt;0[\/latex]. Solving the related equation\u00a0[latex]x^2+2x-15=0[\/latex] gives us the <strong>boundary values<\/strong> which separate solutions from non-solutions on the number line.\u00a0 This equation can be solved by factoring:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2+2x-15&amp;=0\\\\\r\n(x+5)(x-3)&amp;=0\\\\\r\nx+5=0 \\textsf{ or } x-3&amp;=0\\\\\r\nx=-5 \\textsf{ or } x&amp;=3 \\end{align}[\/latex]<\/p>\r\nThus the inequality can switch from being true to false or false to true at these two values. We can test [latex]x[\/latex] values in each of the three number line regions separated by these boundary values to determine if each interval represents solutions or non-solutions.\r\n\r\n[caption id=\"attachment_1885\" align=\"alignnone\" width=\"1194\"]<img class=\"wp-image-1885 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine.png\" alt=\"A number line is shown with blue dots at -5 and 3 to denote boundary values. Test values of -6, 0, and 4 are marked in red.\" width=\"1194\" height=\"105\" \/> Number line for testing values[\/caption]\r\n\r\nThe big dots represent the boundary values and the red Xs represent the test points we will use.\r\n\r\nSubstituting each test value in the original inequality [latex]x^2+2x-15&gt;0[\/latex],\r\n\r\nTest [latex]-6[\/latex]: [latex](-6)^2+2(-6)-15 &gt; 0[\/latex], or [latex]9&gt;0[\/latex] which is TRUE.\r\n\r\nTest [latex]0[\/latex]: [latex](0)^2+2(0)-15 &gt; 0[\/latex], or [latex]-15&gt;0[\/latex] which is FALSE.\r\n\r\nTest [latex]4[\/latex]: [latex](4)^2+2(4)-15 &gt; 0[\/latex], or [latex]9&gt;0[\/latex] which is TRUE.\r\n\r\nThus the solution is [latex](-\\infty, -5) \\cup (3, \\infty)[\/latex], using parentheses at the boundary values because the inequality was strict.\r\n\r\nLet's try a few more examples using this approach.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve the inequality\u00a0[latex]x^2 \\leq 4x + 5[\/latex].\r\n\r\n[reveal-answer q=\"184065\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"184065\"]\r\n\r\nWe should first rewrite the inequality with [latex]0[\/latex] on the right side to prepare to factor.\r\n<p style=\"text-align: center;\">[latex]x^2 - 4x - 5 \\leq 0[\/latex]<\/p>\r\nNow solve for the boundary values by solving the related equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2 - 4x - 5 &amp;= 0\\\\\r\n(x-5)(x+1) &amp;= 0\\\\\r\nx-5=0 \\textsf{ or } x+1&amp;=0\\\\\r\nx=5 \\textsf{ or } x=-1\\end{align}[\/latex]<\/p>\r\nThe boundary values are [latex]x=5[\/latex] and [latex]x=-1[\/latex]. This divides the number line into three distinct intervals. We test a number inside each interval in the original inequality (you could also test it in the inequality solved for 0):\r\n\r\n<img class=\"alignnone wp-image-1900 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine.png\" alt=\"A number line has boundary values marked at -1 and 5, and red test points at -2, 0, and 6.\" width=\"934\" height=\"54\" \/>\r\n\r\nTest [latex]-2[\/latex] from the left interval:\u00a0 [latex](-2)^2 \\leq 4(-2)+5[\/latex], or [latex]4 \\leq -3[\/latex] which is FALSE.\r\n\r\nTest [latex]0[\/latex] from the middle interval:\u00a0 [latex](0)^2 \\leq 4(0)+5[\/latex], or [latex]0 \\leq 5[\/latex] which is TRUE.\r\n\r\nTest [latex]6[\/latex] from the right interval:\u00a0 [latex](6)^2 \\leq 4(6)+5[\/latex], or [latex]36 \\leq 29[\/latex] which is FALSE.\r\n\r\nThus the solution is the interval [latex][-1, 5][\/latex], including the boundary points because equality is allowed in this inequality.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou may have noticed a pattern of the intervals alternating between true and false. Though this is a keen observation, the next example will prove this is not always the case!\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve the inequality\u00a0[latex]2x^2+8x+8 &gt; 0[\/latex].\r\n\r\n[reveal-answer q=\"140424\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"140424\"]\r\n\r\nBegin by solving for boundary values using the related equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x^2+8x+8 &amp;= 0\\\\\r\n2(x^2+4x+4) &amp;= 0\\\\\r\n2(x+2)(x+2) &amp;= 0\\\\\r\nx+2 &amp;= 0\\\\\r\nx&amp;=-2\\end{align}[\/latex]<\/p>\r\nThere is only one boundary value of [latex]x=-2[\/latex] this time because the polynomial to factor was a perfect square trinomial. Our number line has only two intervals to test with one boundary value separating them.\r\n\r\n<img class=\"alignnone wp-image-1901 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine.png\" alt=\"A number line has a boundary value marked at -2 and test points marked at -3 and 0.\" width=\"753\" height=\"71\" \/>\r\n\r\nTest [latex]-3[\/latex] from the left interval:\u00a0 [latex]2(-3)^2 + 8(-3) + 8 &gt; 0[\/latex], or [latex]2 &gt; 0[\/latex] which is TRUE.\r\n\r\nTest [latex]0[\/latex] from the right interval:\u00a0 [latex]2(0)^2 + 8(0) + 8 &gt; 0[\/latex], or [latex]8 &gt; 0[\/latex] which is TRUE.\r\n\r\nBoth intervals consist of solutions this time! The only number which is NOT a solution is the boundary value of [latex]x=-2[\/latex] since the inequality was strict. Thus the solution set is [latex](-\\infty, -2) \\cup (-2, \\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe boundary value and test point method for solving quadratic inequalities will be more generally useful for you in future math courses, where you may solve polynomial or rational inequalities.\r\n<h2>Summary<\/h2>\r\nA quadratic inequality can be solved by two different methods. First make sure [latex]0[\/latex] is on one side of the inequality.\r\n\r\nOne method is to graph the corresponding parabola and determine for which [latex]x[\/latex]-values the parabola is above or below the [latex]x[\/latex]-axis.\r\n\r\nThe other method is to find the endpoints of the solution intervals, called boundary values, by turning the inequality into an equation and solving. Separate the number line into intervals using the boundary values. We can test each resulting interval using a test point to determine whether that interval is in the solution set or not.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve quadratic inequalities, expressing the solution set using interval notation.<\/li>\n<\/ul>\n<\/div>\n<p>A <strong>quadratic inequality<\/strong> is a quadratic equation, but with the equal sign replaced by an inequality sign. An example of a quadratic inequality is<\/p>\n<p style=\"text-align: center;\">[latex]x^2-3x-4\\leq 0[\/latex]<\/p>\n<p>To help solve this, we will consider the quadratic function [latex]f(x)=x^2-3x-4[\/latex]. Solving our inequality above is equivalent to asking which [latex]x[\/latex] values cause the output value of the function [latex]f[\/latex] to be less than or equal to [latex]0[\/latex]. We will use a graph of the function [latex]f[\/latex] to understand this idea. Below is the graph of the function [latex]f[\/latex]. Note that the function factors as [latex]f(x)=(x-4)(x+1)[\/latex], which reveals the two [latex]x[\/latex]-intercepts of [latex](-1,0)[\/latex] and [latex](4, 0)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1878 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-271x300.png\" alt=\"A parabola is facing upwards and has x-intercepts at (-1, 0) and (4, 0)\" width=\"271\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-271x300.png 271w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-65x72.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-225x249.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph-350x387.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph.png 619w\" sizes=\"auto, (max-width: 271px) 100vw, 271px\" \/><\/p>\n<p>Try to use the above graph to solve the inequality [latex]f(x) \\leq 0[\/latex] before revealing the answer below.<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the inequality\u00a0[latex]x^2-3x-4\\leq 0[\/latex] by graphing.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q210097\">Show Solution<\/span><\/p>\n<div id=\"q210097\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solution to the inequality is the portion of the graph of\u00a0[latex]f(x)=x^2-3x-4[\/latex] which has outputs less than or equal to [latex]0[\/latex], which in the graph is the portion of the curve touching or below the [latex]x[\/latex]-axis. We can determine visually that [latex]x[\/latex] values between [latex]-1[\/latex] and [latex]4[\/latex] (and including these two numbers) satisfy this condition. Here is the same graph, but with the portion below the [latex]x[\/latex]-axis colored in green and the portion above the graph dotted:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1879 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-272x300.png\" alt=\"The parabola from the previous graph is marked with the portion below the x-axis highlighted in green\" width=\"272\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-272x300.png 272w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-65x72.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-225x249.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2-350x387.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstGraph2.png 611w\" sizes=\"auto, (max-width: 272px) 100vw, 272px\" \/><\/p>\n<p>The solution is the interval of [latex]x[\/latex]-values corresponding to the region below or touching the [latex]x[\/latex]-axis, which is the interval [latex][-1,4][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can see from the above example that the key was to use the places where the graph <em>crosses<\/em> the [latex]x[\/latex]-axis as the endpoints of our solution interval.<\/p>\n<p>Here is another example to try. Note the small differences in the setup of the problem.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve the inequality\u00a0[latex]x^2-6x+8 > 0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q908945\">Show Solution<\/span><\/p>\n<div id=\"q908945\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function [latex]f(x)=x^2-6x+8[\/latex] factors as [latex]f(x)=(x-4)(x-2)[\/latex] and so the [latex]x[\/latex]-intercepts are at [latex](4,0)[\/latex] and [latex](2,0)[\/latex]. Since the leading coefficient is positive, we draw an upward facing parabola. This gives us enough information to sketch a graph.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1880 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-300x300.png\" alt=\"An upward facing parabola has x-intercepts at (2,0) and (4,0).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph.png 499w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Since we are solving [latex]f(x) > 0[\/latex], we are interested in the portion of the graph that is above the [latex]x[\/latex]-axis this time. Here is the same graph, but with the portion above the graph shaded green and the portion below dotted.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1881 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-300x294.png\" alt=\"The previous parabola is marked with the two portions above the x-axis highlighted in green.\" width=\"300\" height=\"294\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-300x294.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-65x64.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-225x220.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2-350x343.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondGraph2.png 491w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>There are two intervals of [latex]x[\/latex]-values that correspond to portions of the graph above the [latex]x[\/latex]-axis. These intervals consist of [latex]x[\/latex] values less than [latex]2[\/latex] and [latex]x[\/latex] values greater than [latex]4[\/latex], since these are the [latex]x[\/latex]-intercepts of the graph. Since we have strict inequality in this example, the values [latex]x=2[\/latex] and [latex]x=4[\/latex] are not solutions.<\/p>\n<p>The solution is [latex](-\\infty, 2) \\cup (4, \\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using a Number Line<\/h2>\n<p>We can see from the above examples that the [latex]x[\/latex]-intercepts of the function associated with the inequality are crucial to the solving process and end up being the endpoints of the solution intervals. This is because a parabola is a smooth curve and can only pass from above to below the [latex]x[\/latex]-axis by passing through the axis first (resulting in an intercept).<\/p>\n<p>This suggests an alternative approach to solving which does not require graphing the parabola. We can determine the &#8220;boundary values&#8221; where the inequality can pass from being true to false, then use test points to determine which intervals on the number line are true or false. This is similar to the test point approach used in solving systems of inequalities in Chapter 2.<\/p>\n<p>Consider the inequality [latex]x^2+2x-15>0[\/latex]. Solving the related equation\u00a0[latex]x^2+2x-15=0[\/latex] gives us the <strong>boundary values<\/strong> which separate solutions from non-solutions on the number line.\u00a0 This equation can be solved by factoring:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2+2x-15&=0\\\\  (x+5)(x-3)&=0\\\\  x+5=0 \\textsf{ or } x-3&=0\\\\  x=-5 \\textsf{ or } x&=3 \\end{align}[\/latex]<\/p>\n<p>Thus the inequality can switch from being true to false or false to true at these two values. We can test [latex]x[\/latex] values in each of the three number line regions separated by these boundary values to determine if each interval represents solutions or non-solutions.<\/p>\n<div id=\"attachment_1885\" style=\"width: 1204px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1885\" class=\"wp-image-1885 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine.png\" alt=\"A number line is shown with blue dots at -5 and 3 to denote boundary values. Test values of -6, 0, and 4 are marked in red.\" width=\"1194\" height=\"105\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine.png 1194w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-300x26.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-768x68.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-1024x90.png 1024w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-65x6.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-225x20.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6FirstNumberLine-350x31.png 350w\" sizes=\"auto, (max-width: 1194px) 100vw, 1194px\" \/><\/p>\n<p id=\"caption-attachment-1885\" class=\"wp-caption-text\">Number line for testing values<\/p>\n<\/div>\n<p>The big dots represent the boundary values and the red Xs represent the test points we will use.<\/p>\n<p>Substituting each test value in the original inequality [latex]x^2+2x-15>0[\/latex],<\/p>\n<p>Test [latex]-6[\/latex]: [latex](-6)^2+2(-6)-15 > 0[\/latex], or [latex]9>0[\/latex] which is TRUE.<\/p>\n<p>Test [latex]0[\/latex]: [latex](0)^2+2(0)-15 > 0[\/latex], or [latex]-15>0[\/latex] which is FALSE.<\/p>\n<p>Test [latex]4[\/latex]: [latex](4)^2+2(4)-15 > 0[\/latex], or [latex]9>0[\/latex] which is TRUE.<\/p>\n<p>Thus the solution is [latex](-\\infty, -5) \\cup (3, \\infty)[\/latex], using parentheses at the boundary values because the inequality was strict.<\/p>\n<p>Let&#8217;s try a few more examples using this approach.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve the inequality\u00a0[latex]x^2 \\leq 4x + 5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q184065\">Show Solution<\/span><\/p>\n<div id=\"q184065\" class=\"hidden-answer\" style=\"display: none\">\n<p>We should first rewrite the inequality with [latex]0[\/latex] on the right side to prepare to factor.<\/p>\n<p style=\"text-align: center;\">[latex]x^2 - 4x - 5 \\leq 0[\/latex]<\/p>\n<p>Now solve for the boundary values by solving the related equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2 - 4x - 5 &= 0\\\\  (x-5)(x+1) &= 0\\\\  x-5=0 \\textsf{ or } x+1&=0\\\\  x=5 \\textsf{ or } x=-1\\end{align}[\/latex]<\/p>\n<p>The boundary values are [latex]x=5[\/latex] and [latex]x=-1[\/latex]. This divides the number line into three distinct intervals. We test a number inside each interval in the original inequality (you could also test it in the inequality solved for 0):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1900 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine.png\" alt=\"A number line has boundary values marked at -1 and 5, and red test points at -2, 0, and 6.\" width=\"934\" height=\"54\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine.png 934w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine-300x17.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine-768x44.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine-65x4.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine-225x13.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6SecondNumberLine-350x20.png 350w\" sizes=\"auto, (max-width: 934px) 100vw, 934px\" \/><\/p>\n<p>Test [latex]-2[\/latex] from the left interval:\u00a0 [latex](-2)^2 \\leq 4(-2)+5[\/latex], or [latex]4 \\leq -3[\/latex] which is FALSE.<\/p>\n<p>Test [latex]0[\/latex] from the middle interval:\u00a0 [latex](0)^2 \\leq 4(0)+5[\/latex], or [latex]0 \\leq 5[\/latex] which is TRUE.<\/p>\n<p>Test [latex]6[\/latex] from the right interval:\u00a0 [latex](6)^2 \\leq 4(6)+5[\/latex], or [latex]36 \\leq 29[\/latex] which is FALSE.<\/p>\n<p>Thus the solution is the interval [latex][-1, 5][\/latex], including the boundary points because equality is allowed in this inequality.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You may have noticed a pattern of the intervals alternating between true and false. Though this is a keen observation, the next example will prove this is not always the case!<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve the inequality\u00a0[latex]2x^2+8x+8 > 0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q140424\">Show Solution<\/span><\/p>\n<div id=\"q140424\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by solving for boundary values using the related equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x^2+8x+8 &= 0\\\\  2(x^2+4x+4) &= 0\\\\  2(x+2)(x+2) &= 0\\\\  x+2 &= 0\\\\  x&=-2\\end{align}[\/latex]<\/p>\n<p>There is only one boundary value of [latex]x=-2[\/latex] this time because the polynomial to factor was a perfect square trinomial. Our number line has only two intervals to test with one boundary value separating them.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1901 size-full\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine.png\" alt=\"A number line has a boundary value marked at -2 and test points marked at -3 and 0.\" width=\"753\" height=\"71\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine.png 753w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine-300x28.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine-65x6.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine-225x21.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/6.6ThirdNumberLine-350x33.png 350w\" sizes=\"auto, (max-width: 753px) 100vw, 753px\" \/><\/p>\n<p>Test [latex]-3[\/latex] from the left interval:\u00a0 [latex]2(-3)^2 + 8(-3) + 8 > 0[\/latex], or [latex]2 > 0[\/latex] which is TRUE.<\/p>\n<p>Test [latex]0[\/latex] from the right interval:\u00a0 [latex]2(0)^2 + 8(0) + 8 > 0[\/latex], or [latex]8 > 0[\/latex] which is TRUE.<\/p>\n<p>Both intervals consist of solutions this time! The only number which is NOT a solution is the boundary value of [latex]x=-2[\/latex] since the inequality was strict. Thus the solution set is [latex](-\\infty, -2) \\cup (-2, \\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The boundary value and test point method for solving quadratic inequalities will be more generally useful for you in future math courses, where you may solve polynomial or rational inequalities.<\/p>\n<h2>Summary<\/h2>\n<p>A quadratic inequality can be solved by two different methods. First make sure [latex]0[\/latex] is on one side of the inequality.<\/p>\n<p>One method is to graph the corresponding parabola and determine for which [latex]x[\/latex]-values the parabola is above or below the [latex]x[\/latex]-axis.<\/p>\n<p>The other method is to find the endpoints of the solution intervals, called boundary values, by turning the inequality into an equation and solving. Separate the number line into intervals using the boundary values. We can test each resulting interval using a test point to determine whether that interval is in the solution set or not.<\/p>\n","protected":false},"author":773621,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-481","chapter","type-chapter","status-publish","hentry"],"part":199,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/481","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/773621"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/481\/revisions"}],"predecessor-version":[{"id":1915,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/481\/revisions\/1915"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/199"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/481\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=481"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=481"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=481"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=481"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}