{"id":4318,"date":"2013-06-29T03:54:43","date_gmt":"2013-06-29T03:54:43","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4318"},"modified":"2026-03-25T10:47:49","modified_gmt":"2026-03-25T15:47:49","slug":"changing-constants-in-graphs-of-functions-quadratic-equations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/changing-constants-in-graphs-of-functions-quadratic-equations\/","title":{"rendered":"Quadratic Equations Overview"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_t4uZZgsNPjw\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_t4uZZgsNPjw\" data-source-videoID=\"t4uZZgsNPjw\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Quadratic Equations Overview Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Quadratic Equations Overview\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_t4uZZgsNPjw:hover {cursor:pointer;} img#videoThumbnailImage_t4uZZgsNPjw {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1148-quadratic-equations-overview-2.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_t4uZZgsNPjw\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_t4uZZgsNPjw\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Quadratic Equations Overview\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_t4uZZgsNPjw\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_t4uZZgsNPjw\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_t4uZZgsNPjw\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction SDg_Function() {\n  var x = document.getElementById(\"SDg\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"SDg_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"SDg\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Quadratic_Functions\" class=\"smooth-scroll\">Quadratic Functions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Quadratic_Equations_Example\" class=\"smooth-scroll\">Solving Quadratic Equations Example<br \/>\n<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Graphing\" class=\"smooth-scroll\">Graphing<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Completing_the_Square_Method\" class=\"smooth-scroll\">Completing the Square Method<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#The_Quadratic_Formula\" class=\"smooth-scroll\">The Quadratic Formula<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Quadratic_Equation_Practice_Questions\" class=\"smooth-scroll\">Quadratic Equation Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this overview of quadratic equations! Before we dive into how to solve them, let\u2019s first talk about quadratic functions.<\/p>\n<h2><span id=\"Quadratic_Functions\" class=\"m-toc-anchor\"><\/span>Quadratic Functions<\/h2>\n<p>\nWhen we graph quadratic functions, we\u2019ll notice that they can be used to tell all kinds of visual stories, from a daredevil shooting out of a cannon to a satellite dish listening to interstellar signals.<\/p>\n<p>Equations for these functions generally look like this: \\(f(x)=ax^2+bx+c\\) and their graphs form a characteristic shape called a <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/parabolas\/\">parabola<\/a>, which looks something like this one:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/parabola-example-blue.png\" alt=\"\" width=\"390.15\" height=\"396.1\" class=\"aligncenter size-full wp-image-93604\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/parabola-example-blue.png 459w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/parabola-example-blue-295x300.png 295w\" sizes=\"(max-width: 459px) 100vw, 459px\" \/><\/p>\n<p>The different characteristics of quadratic functions that are most commonly analyzed are the vertex (the maximum or minimum point), the <em>x<\/em>-intercepts (the zeros), and the axis of symmetry.<\/p>\n<p>Now that we have a little background, let\u2019s dive further into solving quadratic equations and interpreting the results.<\/p>\n<h2>Solving Quadratic Equations Example<\/h3>\n<p>\nSo what do we mean by \u201csolving\u201d? In this case, one of the things it means is to figure out which values of the variable, if any, make the equation 0. So instead of the function \\(f(x)=ax^2+bx+c\\), we write the related equation: \\(0=ax^2+bx+c\\).<\/p>\n<p>In other words, the solutions to a quadratic equation are the values that make the quadratic function true when \\(f(x)=0\\) or \\(y=0\\).<\/p>\n<p>This is the function \\(f(x)=-x^{2}+2x+8\\):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png\" alt=\"\" width=\"507\" height=\"622\" class=\"aligncenter size-full wp-image-93562\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png 507w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-245x300.png 245w\" sizes=\"auto, (max-width: 507px) 100vw, 507px\" \/><\/p>\n<p>Every point on the graph satisfies this equation.<\/p>\n<p>Solving \\(0=-x^{2}+2x+8\\) gives us two solutions because the function equals 0 only when it crosses the \\(x\\)-axis.<\/p>\n<p>Here\u2019s another way to look at it. To solve a quadratic equation means to find the \\(x\\)-intercepts of the related function, also known as the zeros.<\/p>\n<h3><span id=\"Standard_Form\" class=\"m-toc-anchor\"><\/span>Standard Form<\/h3>\n<p>\nBefore you get started with any solving method, make sure your equation is written in standard form: \\(ax^2+bx+c=0\\). This will prevent errors and can provide insight into choosing the most appropriate method for solving an equation.<\/p>\n<p>If your equation is not in standard form, simply add or subtract terms as needed.<\/p>\n<p>For instance, \\(2x+8=x^{2}\\) is not written in standard form, but by subtracting \\(x^{2}\\) from both sides, we get \\(-x^{2}+2x+8=0\\), which is now in standard form. <\/p>\n<h2><span id=\"Graphing\" class=\"m-toc-anchor\"><\/span>Graphing<\/h2>\n<p>\nThe first method we\u2019ll look at is graphing. If you have access to a graph or can easily create one using the internet or a graphing calculator, this can be a great method to use. Let\u2019s look at an example. Here is the graph of \\(f(x)=-x^{2}+2x+8\\).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png\" alt=\"\" width=\"507\" height=\"622\" class=\"aligncenter size-full wp-image-93562\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png 507w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-245x300.png 245w\" sizes=\"auto, (max-width: 507px) 100vw, 507px\" \/><\/p>\n<p>The solutions to the equation \\(-x^{2}+2x+8=0\\) are \\(x = -2\\) and \\(x = 4\\). Solutions can always be verified by plugging them back in to the original equation:<\/p>\n<div class=\"examplesentence\"><strong>Solution 1<\/strong><br \/>\n\\(-(-2)2 + 2(-2) + 8 = 0\\)<br \/>\n\\(-4 \u2013 4 = 0\\)<br \/>\n\\(0 = 0\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\"><strong>Solution 2<\/strong><br \/>\n\\(-(4)2 + 2(4) + 8 = 0\\)<br \/>\n\\(-16 + 8 + 8 =0\\)<br \/>\n\\(0 = 0\\)<\/div>\n<p>\n&nbsp;<br \/>\nSometimes, quadratic equations can be simplified and solved by factoring.<\/p>\n<p>Looking at our sample equation, we have \\(-x^{2}+2x+8=0\\).<\/p>\n<p>The left side can be factored into \\((-x+4)(x+2)=0\\).<\/p>\n<h3><span id=\"Zero_Product_Property\" class=\"m-toc-anchor\"><\/span>Zero Product Property<\/h3>\n<p>\nNow let\u2019s pause. There\u2019s a property called the zero product property that says that if there are two numbers whose product is 0, one of the numbers must be 0. In more mathematical terms: If \\(a\\) and \\(b\\) are real numbers and \\(ab=0\\), then \\(a=0\\) or \\(b=0\\).<\/p>\n<p>Makes sense, right? Back to our equation, \\((-x+4)(x+2)=0\\).<\/p>\n<p>This means that \\(-x+4=0\\) or \\(x+2=0\\). So we solve both!<\/p>\n<p>We&#8217;re going to subtract 4 from both sides, which gives us \\(-x=-4\\). Then divide by -1 to get \\(x=4\\).<\/p>\n<p>For this equation, we subtract 2 from both sides, and we&#8217;re left with \\(x=-2\\).<\/p>\n<p>So our zeroes are \\(x=4\\) and \\(x=-2\\).<\/p>\n<p>These match what we found on our graph.<\/p>\n<h2><span id=\"Completing_the_Square_Method\" class=\"m-toc-anchor\"><\/span>Completing the Square Method<\/h2>\n<p>\nFactoring is nice, but not every equation can be factored easily. Luckily, there are a couple more methods we can use, including the completing the square method.<\/p>\n<p>The idea behind this method is to create a quadratic equation that contains a perfect square and then take the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/roots\/\">square root<\/a> to find the solutions.<\/p>\n<p>Here\u2019s our equation again: \\(-x^{2}+2x+8=0\\). To complete the square, the equation needs to not only be in standard form, but the coefficient on the squared term needs to be positive instead of negative. So, we divide all terms by -1 to get ready. This gives us \\(x^{2}-2x-8=0\\).<\/p>\n<p>First, move the constant term to the other side by itself. For this, we\u2019ll add 8 to both sides. This gives us \\(x^{2}-2x=8\\).<\/p>\n<p>Now we need to add a number that creates a perfect square on the left-hand side. Let\u2019s consider some perfect squares:<\/p>\n<table class=\"ATable\" style=\"margin: auto;\">\n<thead>\n<tr>\n<th><strong>Factored Form<\/strong><\/th>\n<th><strong>Expanded Form<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height:60px\">\n<td>\\((x+2)^2\\)<\/td>\n<td>\\(x^2+\\) <span style=\"color:#e2ba12\">\\(4\\)<\/span> \\(x\\) \\(+\\) <span style=\"color:#005AB5\">\\(4\\)<\/span><\/td>\n<\/tr>\n<tr style=\"height:60px\">\n<td>\\((x-3)^2\\)<\/td>\n<td>\\(x^2-\\) <span style=\"color:#e2ba12\">\\(6\\)<\/span> \\(x\\) \\(+\\) <span style=\"color:#005AB5\">\\(9\\)<\/span><\/td>\n<\/tr>\n<tr style=\"height:60px\">\n<td>\\((x+4)^2\\)<\/td>\n<td>\\(x^2+\\) <span style=\"color:#e2ba12\">\\(8\\)<\/span> \\(x\\) \\(+\\) <span style=\"color:#005AB5\">\\(16\\)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nTo complete the square in our equation, we&#8217;ll add a number to both sides of our equation. In the examples, notice that number is \\((\\frac{b}{2})^{2}\\).<\/p>\n<div class=\"examplesentence\">\\(4=(\\frac{4}{2})^{2}\\)<br \/>\n\\(9=(\\frac{-6}{2})^{2}\\)<br \/>\nand<br \/>\n\\(16=(\\frac{8}{2})^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn our equation, we\u2019ll add  \\((\\frac{-2}{2})^{2}=1\\), since our \\(b = -2\\). Remember, an equation only stays equal if you change both sides in the same way, so since we&#8217;re gonna add 1 to our left side, we also need to be sure to add 1 to the right side: \\(x^{2} \u2013 2x + 1 =9\\).<\/p>\n<p>Now the left side can be factored: \\((x-1)^{2}=9\\)<\/p>\n<p>Since we have a perfect square, we take the square root of both sides. On the right-hand side, we need to account for both positive and negative square roots.<\/p>\n<div class=\"examplesentence\">\\( \\sqrt{(x-1)^{2}}=\u00b1\\sqrt{9}\\)<br \/>\n\\(x-1=\u00b13\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we\u2019ll split into separate mini-equations: \\(x-1=3\\) and \\(x-1=-3\\). We&#8217;re going to solve for \\(x\\) over here (\\(x-1=3\\)) by adding 1 to both sides, which gives us \\(x=4\\). And here (\\(x-1=-3\\)) we&#8217;re going to do the same; add 1 to both sides, which gives us \\(x=-2\\).<\/p>\n<p>We get the same solutions as before.<\/p>\n<p>Completing the square can be used with any quadratic equation, but if you start with a perfect square and don\u2019t realize it, the equation will look the same after you use this method as it did when you started.<\/p>\n<h2><span id=\"The_Quadratic_Formula\" class=\"m-toc-anchor\"><\/span>The Quadratic Formula<\/h2>\n<p>\nThere is one last method for us to examine. The quadratic formula, as you\u2019ll see, involves the most algebra, but it has two distinct characteristics: <\/p>\n<ol>\n<li>It can truly be used with any quadratic equation.<\/li>\n<li>Part of it can be used to determine how many solutions to expect beforehand.<\/li>\n<\/ol>\n<p>Here it is in all its glory: <span style=\"font-size:130%\">\\(x=\\frac{-b\\pm \\sqrt{b^{2}-4ac}}{2a}\\)<\/span><\/p>\n<p>The piece of the quadratic formula under the radical, the square root sign, is called the <strong>discriminant<\/strong>. Calculating its value helps us know how many solutions to expect.<\/p>\n<p>Let\u2019s take a minute to consider how many times a quadratic function can possibly intersect the \\(x\\)-axis:<\/p>\n<table class=\"ATable\" style=\"margin: auto;\">\n<tbody>\n<tr>\n<td><strong>This example shows twice<\/strong><\/td>\n<\/tr>\n<tr>\n<td><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png\" alt=\"\" width=\"405.6\" height=\"497.6\" class=\"aligncenter size-full wp-image-93562\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola.png 507w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-245x300.png 245w\" sizes=\"(max-width: 507px) 100vw, 507px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<\/p>\n<table class=\"ATable\" style=\"margin: auto;\">\n<tbody>\n<tr>\n<td><strong>Sometimes, it&#8217;s only the vertex<\/strong><\/td>\n<\/tr>\n<tr>\n<td><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-2.4.png\" alt=\"\" width=\"405\" height=\"570.0519\" class=\"aligncenter size-full wp-image-93580\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-2.4.png 454w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-2.4-213x300.png 213w\" sizes=\"(max-width: 454px) 100vw, 454px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<\/p>\n<table class=\"ATable\" style=\"margin: auto;\">\n<tbody>\n<tr>\n<td><strong>Sometimes, it doesn&#8217;t intersect<\/strong><\/td>\n<\/tr>\n<tr>\n<td><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-3.png\" alt=\"\" width=\"405\" height=\"570.0519\" class=\"aligncenter size-full wp-image-93583\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-3.png 454w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/function-parabola-3-213x300.png 213w\" sizes=\"(max-width: 454px) 100vw, 454px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nBecause of the shape, these are our only choices. In order for a graph like this to intersect the \\(x\\)-axis more than two times, it would need to change direction again and thus, no longer be a parabola.<\/p>\n<p>Now let\u2019s consider square roots in general. If the square root of a number is greater than 0, there are two real solutions; taking the square root of a positive number yields positive and negative roots. If the square root is 0, there&#8217;s one real solution: 0. Taking \\(\\sqrt{0}=0\\). If the square root of a number is less than 0, there are zero real solutions. We can\u2019t take the square root of a negative number and end up with a real number.<\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 100%;\">\n<thead>\n<tr style=\"height:50px\">\n<th style=\"vertical-align: middle;\">If \\(\\sqrt{\\text{number > }0}\\)<\/th>\n<th style=\"width: 30%; vertical-align: middle;\">If \\(\\sqrt{0}\\)<\/th>\n<th style=\"vertical-align: middle;\">If \\(\\sqrt{\\text{number &#60; }0}\\)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height:50px\">\n<td>2 real solutions<\/td>\n<td>1 real solution = 0<\/td>\n<td>0 real solutions<\/td>\n<\/tr>\n<tr style=\"height:50px\">\n<td>Taking the square root of a positive number yields \u00b1 roots<\/td>\n<td>Taking the square root<br \/>\nof 0 is 0<\/td>\n<td>Taking the square of a negative number does not result in a real number<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nSo what does this have to do with the discriminant?<\/p>\n<ol>\n<li style=\"margin-bottom: 10px;\">If \\(b^{2}-4ac\\) > \\(0\\), then there are two real solutions (two \\(x\\)-intercepts).<\/li>\n<li style=\"margin-bottom: 10px;\">2. If \\(b^{2}-4ac=0\\), there is one real solution (the vertex of the parabola is the \\(x\\)-intercept).<\/li>\n<li>And if \\(b^{2}-4ac\\) < \\(0\\), there are no real solutions (no \\(x\\)-intercepts).<\/li>\n<\/ol>\n<p>Using our ongoing example, \\(-x^{2}+2x+8=0\\): \\(a = -1\\), \\( b = 2\\), and \\(c = 8\\).<\/p>\n<p>The discriminant, \\(D\\), is \\(b^{2}-4ac\\), which is:<\/p>\n<div class=\"examplesentence\">\\((2)^{2}-4(-1)(8)\\)<br \/>\n\\(=4+32\\)<br \/>\n\\(=36\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince \\(36\\) > \\(0\\), we know that there are two real solutions to our equation, as expected.<\/p>\n<p>What\u2019s the point of calculating the discriminant? If you get in the habit of checking it every time, a few things happen:<\/p>\n<ul>\n<li>If there are no solutions to your equation, you\u2019ll know right away and can stop<\/li>\n<li>If there are one or two solutions, the discriminant will serve as a check, regardless of the method used<\/li>\n<li>If you use the quadratic formula, a chunk of it is already calculated<\/li>\n<\/ul>\n<p>Speaking of the quadratic formula, let\u2019s try it out!<\/p>\n<p>Again, \\(a = -1\\), \\( b = 2\\), and \\(c = 8\\). Plus, we already know the discriminant. So we have \\(-b\\), which is \\(-2\\pm \\sqrt{\\text{discriminant}}\\), which we found was 36 (so \\(\\sqrt{36}\\)) all over \\(2a\\), which is -1: \\(\\frac{-2\\pm \\sqrt{36}}{2(-1)}\\).<\/p>\n<p>This is equal to \\(\\frac{-2\\pm 6}{-2}\\).<\/p>\n<p>Now we\u2019re going to split it into two equations. We have \\(\\frac{-2+ 6}{-2}\\) and then \\(\\frac{-2- 6}{-2}\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{-2+ 6}{-2}\\) gives you \\(\\frac{-2+ 6}{-2}=\\frac{4}{-2}=-2\\).<br \/>\n\\(\\frac{-2- 6}{-2}\\) gives us \\(\\frac{-2- 6}{-2}=\\frac{-8}{-2}=4\\).<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(x=-2\\) and \\(x=4\\).<\/p>\n<p>We\u2019ve seen all four quadratic equation solving methods in action. Now let\u2019s put some context around our practice equation.<\/p>\n<p>Suppose our function \\(f(x)-x^{2}+2x+8\\) represents the height of a rock from the ground (in meters) you throw off of a small cliff as a function of time (in seconds). We want to know how long it takes for the rock to hit the ground.<\/p>\n<p>In order to do that, we would solve the equation \\(-x^{2} +2x + 8 = 0\\), which we have already done many times over. Remember, the 0 represents 0 height. In function terms, this means \u201con the \\(x\\)-axis.\u201d In this situation, that means \u201con the ground.\u201d<\/p>\n<p>We know our mathematical solutions are \\(x=-2\\) and \\(x=4\\). In this situation, the solution of -2 doesn\u2019t make sense. What does it mean? 2 seconds ago? Travel back in time 2 seconds? We would state that the rock landed after 4 seconds.<\/p>\n<p>I hope this video helped you understand quadratic equations and how they work! Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Quadratic_Equation_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Quadratic Equation Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhat is the solution to the following quadratic equation?<\/p>\n<div class=\"yellow-math-quote\">\\(x^2+7x+15=5\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x=2\\) and \\(x=5\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">\\(x=-2\\) and \\(x=-5\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(x=1\\) and \\(x=10\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(x=-1\\) and \\(x=-10\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The first step to solving a quadratic equation is to set the function equal to zero. We can subtract 5 from both sides to make the equation \\(x^2+7x+10=0\\). Once it is in this form, we should always check the discriminant, \\(b^2-4ac\\), to determine the number of real solutions to the quadratic equation. There are two in this case because \\(7^2-4(1)(10)=9\\), which is greater than zero.<\/p>\n<p>The second step is to see if we can solve this by factoring. The factors of 10 are 5 and 2, and 10 and 1. It&#8217;s easy to see that the combination of 5 and 2 add to 7, which is the coefficient of our middle term.<\/p>\n<p>The factored form of this equation is \\((x+2)(x+5)=0\\). We can use the zero product property to set each factor equal to zero and solve for \\(x\\), which gives us \\(x=-2\\) and \\(x=-5\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich equation shows the following quadratic equation after completing the square? <\/p>\n<div class=\"yellow-math-quote\">\\(x^2-18x+8=-9\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\((x+3)^2=26\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\((x-3)^2=26\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((x+9)^2=64\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\((x-9)^2=64\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We must first check the discriminant, \\(b^2-4ac\\), to see if there are any solutions and the number of solutions to the quadratic equation.<\/p>\n<p>First, we will set the equation equal to zero by adding 9 to both sides to get \\(x^2-18x+17=0\\). The discriminant is \\((-18)^2-4(1)(17)=256\\), which is greater than zero; therefore, this quadratic equation has two real solutions.<\/p>\n<p>When solving a quadratic equation using the completing the square method, the first step is to move all the constants to the right side of the equation:<\/p>\n<p style=\"text-align: center\">\\(x^2-18x=-17\\)<\/p>\n<p>Then, we take the coefficient of \\(x\\) and divide it by two and square it:<\/p>\n<p style=\"text-align: center\">\\((\\frac{-18}{2})^2=81\\)<\/p>\n<p>This becomes the third term in our quadratic equation, but you must add it to both sides to maintain the equality of the equation and get \\(x^2-18x+81=-17+81\\).<\/p>\n<p>Simplify the equation \\(x^2-18x+81=64\\). Then, factor the quadratic equation \\((x-9)(x-9)=64\\), which can be written as \\((x-9)^2=64\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhat is the solution to the following quadratic equation?<\/p>\n<div class=\"yellow-math-quote\">\\(-8x^2-3x+22=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x=-1.856\\) and \\(x=-1.481\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(x=1.856\\) and \\(x=1.481\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-3\">\\(x=-1.856\\) and \\(x=1.481\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(x=1.856\\) and \\(x=-1.481\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We will start by checking the discriminant, \\(b^2-4ac\\), to determine the number of solutions for the quadratic equation.<\/p>\n<p style=\"text-align: center\">\\((-3)^2-4(-8)(22)=713\\)<\/p>\n<p>Therefore, there are two real solutions.<\/p>\n<p>Since there is a leading coefficient that does not divide equally into the other terms we will use the quadratic formula to solve the equation. Start by substituting the values into the formula:<\/p>\n<p style=\"text-align: center\">\\(x=\\dfrac{-(-3)\\pm\\sqrt{(-3)^2-4(-8)(22)}}{2(-8)}\\)<\/p>\n<p>Simplify the equation, then solve for each value.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(x=\\dfrac{3\\pm\\sqrt{713}}{-16}\\)<br \/>\n\\(x=-1.856\\) and \\(x=1.481\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSmith throws a ball into the air. The height (\\(h\\)) of the ball in feet above the ground after \\(t\\) seconds can be shown using the function below. How many seconds is the ball in the air?<\/p>\n<div class=\"yellow-math-quote\">\\(h(t)=-2x^2+7x+12\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">1.26 seconds<\/div><div class=\"PQ\"  id=\"PQ-4-2\">3.50 seconds<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">4.76 seconds<\/div><div class=\"PQ\"  id=\"PQ-4-4\">6.02 seconds<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>To find the number of seconds the ball is in the air we will solve for \\(t\\) using the quadratic formula.<\/p>\n<p>We will begin by substituting the values for \\(a\\), \\(b\\), and \\(c\\) into the formula:<\/p>\n<p style=\"text-align: center\">\\(x=\\dfrac{-7\\pm\\sqrt{(7)^2-4(-2)(12)}}{2(-2)}\\)<\/p>\n<p>Simplify the equation, then solve for each.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(x=\\dfrac{-7\\pm\\sqrt{145}}{-4}\\)<br \/>\n\\(x=-1.26\\) and \\(x=4.76\\)<\/p>\n<p>Since the question is asking for the number of seconds and we cannot have a negative number of seconds, we can conclude that the solution is only \\(x=4.76\\) seconds.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhat is the total number of real solutions for the following quadratic equation?<\/p>\n<div class=\"yellow-math-quote\">\\(9x^2+12x+4=0\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">One real solution<\/div><div class=\"PQ\"  id=\"PQ-5-2\">Two real solutions<\/div><div class=\"PQ\"  id=\"PQ-5-3\">No real solution<\/div><div class=\"PQ\"  id=\"PQ-5-4\">Infinite real solutions<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>To determine the number of real solutions, we use the discriminant, \\(b^2-4ac\\), which simplifies to 0 because \\((12)^2-4(9)(4)=144-144=0\\).<\/p>\n<p>Therefore, there is one real solution to this quadratic equation.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-ii\/\">Return to Algebra II Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra II Videos<\/p>\n","protected":false},"author":1,"featured_media":100264,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4318","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-how-graphs-change-when-the-constants-change-videos","7":"page_category-math-advertising-group","8":"page_category-video-pages-for-study-course-sidebar-ad","9":"page_type-video","10":"content_type-practice-questions","11":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4318","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=4318"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4318\/revisions"}],"predecessor-version":[{"id":278944,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4318\/revisions\/278944"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100264"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4318"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}