{"id":88,"date":"2017-08-23T07:45:13","date_gmt":"2017-08-23T07:45:13","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=88"},"modified":"2017-09-18T15:50:30","modified_gmt":"2017-09-18T15:50:30","slug":"solving-and-graphing-quadratic-equations","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/solving-and-graphing-quadratic-equations\/","title":{"rendered":"Solving and Graphing Quadratic Equations"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/simplifying-rational-polynomials\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/algebra-functions-ii\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/binomial-expansion\">Next Lesson \u27a1<\/a><\/div>\n<p><!-- CONTENT BEGINS HERE --><\/p>\n<h1 id=\"title\">Solving and Graphing Quadratic Equations<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, you will study two different forms of quadratic equations: standard form and vertex form. Also, you\u00a0will study how to find the roots of quadratics both algebraically and graphically. You will be introduced to the\u00a0concept of symmetry in graphs of parabolas.<\/p>\n<h4>Previously Covered:<\/h4>\n<ul>\n<li><em><strong>Rational\u00a0polynomials<\/strong><\/em> can be easily simplified when they are\u00a0factored (if possible) and common terms are eliminated. Always\u00a0look for a <em>GCF<\/em> before factoring.<\/li>\n<li>The operations addition,\u00a0subtraction, multiplication, and\u00a0division can be polynomials to combine multiple\u00a0rational polynomials into one final solution.<\/li>\n<li>Rational functions are undefined at their\u00a0<em><strong>asymptotes<\/strong><\/em>.<\/li>\n<\/ul>\n<section>\n<h3>What is the standard form of a\u00a0quadratic function?<\/h3>\n<p>The standard form of a\u00a0quadratic function is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image003.gif\" width=\"122\" height=\"17\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0where <em>a<\/em>,\u00a0<em>b<\/em>,\u00a0and <em>c\u00a0<\/em>are real numbers, and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image006.gif\" width=\"31\" height=\"11\" name=\"graphics4\" align=\"absmiddle\" border=\"0\" \/><em>.\u00a0<\/em>Each term in the\u00a0function has a special purpose:<\/p>\n<p style=\"text-decoration: none;\" align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image009.gif\" width=\"20\" height=\"14\" name=\"graphics5\" align=\"absmiddle\" border=\"0\" \/>\u00a0is the quadratic term.<\/p>\n<p style=\"text-decoration: none;\" align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image012.gif\" width=\"15\" height=\"11\" name=\"graphics6\" align=\"absmiddle\" border=\"0\" \/>\u00a0is the linear term.<\/p>\n<p align=\"CENTER\"><em>c\u00a0<\/em>is the constant term.<\/p>\n<p style=\"text-decoration: none;\">The coefficient of the quadratic term,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image015.gif\" width=\"7\" height=\"7\" name=\"graphics7\" align=\"absmiddle\" border=\"0\" \/>,\u00a0determines how wide or narrow the graphs are, and whether the<br \/>\ngraph turns upward or downward.<\/p>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<ul>\n<li class=\"notebox_text\">A positive quadratic coefficient causes the ends of the parabola to point\u00a0upward.<\/li>\n<li class=\"notebox_text\">A negative quadratic coefficient causes the ends of the parabola\u00a0to point downward.<\/li>\n<li class=\"notebox_text\">The greater the quadratic coefficient, the narrower the\u00a0parabola.<\/li>\n<li class=\"notebox_text\">The lesser the quadratic coefficient, the wider the parabola.<\/li>\n<\/ul>\n<\/div>\n<p>In this graph, coefficient <em>a\u00a0<\/em>is positive and large. Therefore the parabola is narrow and points\u00a0upward.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20001.JPG\" alt=\"Narrow parabola opening upward\" width=\"134\" height=\"123\" name=\"graphics8\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p>In this graph, coefficient\u00a0<em>a\u00a0<\/em>is smaller. Therefore, the parabola is wider.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20002.JPG\" alt=\"Wider parabola opening upward\" width=\"133\" height=\"124\" name=\"graphics9\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p>Here, coefficient <em>a\u00a0<\/em>is negative, therefore the endpoints of the parabola point\u00a0downward.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20003.JPG\" alt=\"Parabola opening downward\" width=\"133\" height=\"124\" name=\"graphics10\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p>The linear-term coefficient\u00a0<em>b\u00a0<\/em>shifts the axis of symmetry away from the <em>y<\/em>-axis.\u00a0The direction of shift depends on the sign of the quadratic\u00a0coefficient and the sign of the linear coefficient.<\/p>\n<p style=\"text-decoration: none;\">The axis of symmetry shifts to\u00a0the right if the equation has:<\/p>\n<ul>\n<li>positive\u00a0<em>a\u00a0<\/em>and negative <em>b\u00a0<\/em>coefficients, or<\/li>\n<li>negative <em>a\u00a0<\/em>and positive <em>b\u00a0<\/em>coefficients.<\/li>\n<\/ul>\n<p style=\"text-decoration: none;\">The axis of symmetry shifts to\u00a0the left if the equation has:<\/p>\n<ul>\n<li>positive\u00a0<em>a<\/em> and positive <em>b\u00a0<\/em>coefficients, or<\/li>\n<li>negative <em>a\u00a0<\/em>and negative <em>b\u00a0<\/em>coefficients.<\/li>\n<\/ul>\n<p>The constant term <em>c\u00a0<\/em>affects the <em>y<\/em>-intercept.\u00a0The greater the number, the higher the intercept point on the <em>y\u00a0<\/em>-axis.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>Which graph best matches the quadratic function\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image018.gif\" width=\"123\" height=\"17\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>?<\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20004.JPG\" alt=\" Parabola with linear coefficient -2 that opens upward\" width=\"133\" height=\"124\" name=\"graphics12\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20005.JPG\" alt=\"Parabola shifted left that opens upward\" width=\"134\" height=\"121\" name=\"graphics13\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.2%20Art%20006.JPG\" alt=\" Parabola shifted right, opening down\" width=\"133\" height=\"125\" name=\"graphics14\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math%20Mod%203.1%20Art%20007.JPG\" alt=\" Parabola shifted left, opening down\" width=\"134\" height=\"124\" name=\"graphics15\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct choice is A. The quadratic coefficient is +3,\u00a0therefore the endpoints of the graph will turn upward. This\u00a0eliminates choices C and D. The\u00a0<em>y<\/em>-intercept\u00a0is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p2_clip_image021.gif\" width=\"16\" height=\"11\" name=\"graphics16\" align=\"absmiddle\" border=\"0\" \/>.\u00a0A linear coefficient of \u20132 paired with a positive quadratic\u00a0coefficient indicates the axis if symmetry will shift to the\u00a0right.<\/p>\n<\/div>\n<\/section>\n<h3>What is the vertex form of a quadratic equation?<\/h3>\n<p>There are times when the minimum or maximum point of a\u00a0parabola, called the <abbr title=\" minimum or maximum point of a parabola\">vertex<\/abbr>,\u00a0is not located at the origin. When this is the case, we can write\u00a0the quadratic function in vertex form, in order to easily identify\u00a0the coordinate location of the vertex, and thus the axis of\u00a0symmetry.\u00a0Vertex form is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image003.gif\" width=\"108\" height=\"17\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p>The variables (<em>h<\/em>,\u00a0<em>k<\/em>)\u00a0are the coordinate location that determines the position of the\u00a0vertex. The quadratic coefficient <em>a\u00a0<\/em>determines the direction of the end points of the parabola.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>What is the equation of the quadratic function modeled in the\u00a0following graph?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math-Mod-3.2-Art-007.gif\" width=\"398\" height=\"257\" name=\"graphics4\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<ol>\n<li><a name=\"one\"><\/a><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image006.gif\" width=\"110\" height=\"34\" name=\"graphics5\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image009.gif\" width=\"121\" height=\"34\" name=\"graphics6\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image012.gif\" width=\"110\" height=\"34\" name=\"graphics7\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image015.gif\" width=\"121\" height=\"34\" name=\"graphics8\" align=\"absmiddle\" border=\"0\" \/><\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The vertex is located at\u00a0(\u20132, 3), and a second point on the graph is located at\u00a0(\u20134, \u20134). Therefore, <em>h<br \/>\n= <\/em>\u20132,\u00a0<em>k <\/em>= 3,\u00a0<em>x <\/em>= \u20134, and\u00a0<em>y <\/em>= \u20134<em>.\u00a0<\/em>You can now solve for <em>a\u00a0<\/em>using the vertex\u00a0form, as shown below.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image018.gif\" width=\"140\" height=\"135\" name=\"graphics9\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p>The quadratic function is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image021.gif\" width=\"119\" height=\"34\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0The correct choice is B.<\/p>\n<\/div>\n<\/section>\n<h3>How do we solve quadratic functions by algebraic methods?<\/h3>\n<p>When we solve quadratic\u00a0functions, we are actually finding the roots of the equation. A\u00a0<em>root\u00a0<\/em>is a point where the equation is equal to zero (in algebraic\u00a0terms), or where the graph crosses the <em>x<\/em>-axis\u00a0(in graphical terms).<\/p>\n<h4><em><strong>Method one: Solve by factoring. <\/strong><\/em><\/h4>\n<p>Factor to solve the quadratic equation 14<em>x<\/em><sup>2<\/sup> +\u00a07<em>x<\/em> = 0<em>.<\/em><\/p>\n<p><strong>Step 1:<\/strong> Factor out a GCF of 7<em>x<\/em>, and\u00a0rewrite the equation as 7<em>x<\/em>(2<em>x <\/em>+ 1) = 0<em>.<\/em><\/p>\n<p><strong>Step 2:<\/strong> Since the first term, 7<em>x<\/em>, is\u00a0multiplied by the second binomial, 2<em>x <\/em>+ 1, we have to\u00a0determine what numbers are required to make the equation equal\u00a0zero. This gives us two equations.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image038.gif\" width=\"60\" height=\"35\" name=\"graphics11\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 3:<\/strong> The\u00a0roots are <em>x = <\/em>0\u00a0and <em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image041.gif\" width=\"45\" height=\"34\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0<\/em>.<\/p>\n<p>For example, use factoring to solve the quadratic equation\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image044.gif\" width=\"92\" height=\"14\" name=\"graphics13\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<p><strong>Step 1:<\/strong> The quadratic factors into\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image047.gif\" width=\"123\" height=\"15\" name=\"graphics14\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<p><strong>Step 2:<\/strong> Set each binomial term equal to zero,\u00a0and solve.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image050.gif\" width=\"62\" height=\"107\" name=\"graphics15\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 3:<\/strong> The roots are\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image053.gif\" width=\"39\" height=\"11\" name=\"graphics16\" align=\"absmiddle\" border=\"0\" \/>\u00a0and\u00a0<em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p3_clip_image056.gif\" width=\"40\" height=\"11\" name=\"graphics17\" align=\"absmiddle\" border=\"0\" \/>.<\/em><\/p>\n<h4>Method Two: Solve by completing the square<\/h4>\n<p class=\"lesson_text\">You are able to solve quadratic equations by completing the\u00a0square because if you square a binomial, the result is a <abbr title=\"the trinomial that results from the square of a binomial\">perfect\u00a0square trinomial<\/abbr>.<\/p>\n<table width=\"75%\">\n<thead>\n<tr>\n<th>Binomial<\/th>\n<th>Perfect Square Trinomial<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image003.gif\" width=\"45\" height=\"17\" name=\"graphics3\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image006.gif\" width=\"68\" height=\"14\" name=\"graphics4\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<\/tr>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image009.gif\" width=\"45\" height=\"17\" name=\"graphics5\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image012.gif\" width=\"83\" height=\"14\" name=\"graphics6\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<\/tr>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image015.gif\" width=\"55\" height=\"17\" name=\"graphics7\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/s4_p4_html_mae77a54.gif\" width=\"96\" height=\"21\" name=\"graphics8\" align=\"absmiddle\" border=\"0\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p class=\"lesson_text\">The first step is to find the perfect square trinomial, and\u00a0then write the trinomial as a binomial squared, using the\u00a0following formula.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image021.gif\" width=\"165\" height=\"43\" name=\"graphics9\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p>For example, solve\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image024.gif\" width=\"101\" height=\"14\" name=\"graphics10\" align=\"absmiddle\" border=\"0\" \/>\u00a0by completing the square.<\/p>\n<p><strong>Step 1:<\/strong> Rewrite the quadratic equation with\u00a0the terms containing <em>x\u00a0<\/em>on the left side\u00a0of the equal sign, and the constant on the right side of the equal\u00a0sign.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image027.gif\" width=\"78\" height=\"14\" name=\"graphics11\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> Divide\u00a0the coefficient <em>b\u00a0<\/em>by 2, square the result, and add this to both sides of the\u00a0equation.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image030.gif\" width=\"192\" height=\"43\" name=\"graphics12\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 3:<\/strong> Simplify. Notice that the left side is\u00a0a perfect square trinomial.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image033.gif\" width=\"101\" height=\"43\" name=\"graphics13\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 4:<\/strong> Take the square root of both sides,\u00a0and solve for the roots.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p4_clip_image036.gif\" width=\"105\" height=\"85\" name=\"graphics14\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<h4>Method Three: Solve using the quadratic formula.<\/h4>\n<p class=\"lesson_text\">The quadratic\u00a0formula is a straightforward method that always\u00a0produces a solution, when other methods such as graphing,\u00a0factoring, and completing the square require too much effort or\u00a0are not easily done. The formula is<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image003.gif\" width=\"100\" height=\"41\" name=\"graphics3\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">A part of this formula, called the <abbr title=\" the section of the quadratic formula beneath the radical\">discriminate<\/abbr>,\u00a0quickly provides valuable information about the roots. The\u00a0discriminate is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image006.gif\" width=\"65\" height=\"20\" name=\"graphics4\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<p>If <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image008.gif\" width=\"65\" height=\"20\" name=\"graphics5\" align=\"absmiddle\" border=\"0\" \/> is:<\/p>\n<ul>\n<li class=\"notebox_text\">positive, there will be two real roots and no imaginary roots.<\/li>\n<li class=\"notebox_text\">negative, there will be two imaginary roots and no real roots.<\/li>\n<li class=\"notebox_text\">zero, there will be one real root.<\/li>\n<\/ul>\n<\/div>\n<p>For example, solve the equation\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image011.gif\" width=\"97\" height=\"14\" name=\"graphics6\" align=\"absmiddle\" border=\"0\" \/><em>.<br \/>\n<\/em><\/p>\n<p><strong>Step 1:<\/strong> Identify\u00a0the coefficients: <em>a\u00a0<\/em>= 3,\u00a0<em>b <\/em>= 2, and <em>c\u00a0<\/em>= 1. Therefore,\u00a0the discriminate is:<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image014.gif\" width=\"205\" height=\"20\" name=\"graphics7\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<p>A negative under the radical means the solution will have\u00a0imaginary roots.<\/p>\n<p><strong>Step 2:<\/strong> Apply the quadratic formula:.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image017.gif\" width=\"343\" height=\"46\" name=\"graphics8\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 3:<\/strong> Simplify further, and write the\u00a0solution in imaginary form.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image020.gif\" width=\"89\" height=\"45\" name=\"graphics9\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<p>Remember, for an imaginary number, <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image023.gif\" width=\"48\" height=\"19\" name=\"graphics10\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<\/div>\n<h4>Method Four: Solve quadratic equations by graphing<\/h4>\n<p>Graphing provides a visual way to easily identify the roots,\u00a0and see how the equation behaves over a given domain and range.<\/p>\n<p>For example, solve\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image026.gif\" width=\"61\" height=\"14\" name=\"graphics11\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<p><strong>Step 1:<\/strong> Draw the graph, and identify where the\u00a0parabola crosses the <em>x-<\/em>axis.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/Math-Mod3.2-Art-008.gif\" width=\"417\" height=\"233\" name=\"graphics12\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<p><strong>Step 2:<\/strong> The\u00a0<em>x<\/em>-axis\u00a0crossing points are the solution to the equation, where <em>y\u00a0<\/em>= 0. Therefore, the roots are <em>x\u00a0<\/em>= 2, and <em>x\u00a0= <\/em>\u20132.<\/p>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<p>We could also write the solution as <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image029.gif\" width=\"109\" height=\"14\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/>. If we multiply the binomials, the result is the original equation, <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/3\/images\/s4_p5_clip_image031.gif\" width=\"61\" height=\"14\" name=\"graphics14\" align=\"absmiddle\" border=\"0\" \/><\/p>\n<\/div>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/simplifying-rational-polynomials\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/algebra-functions-ii\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/binomial-expansion\">Next Lesson \u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Solving and Graphing Quadratic Equations Objective In this lesson, you will study two different forms of quadratic equations: standard form and vertex form. Also, you\u00a0will study how to find the roots of quadratics both algebraically and graphically. You will be introduced to the\u00a0concept of symmetry in graphs of parabolas. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-88","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/88","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=88"}],"version-history":[{"count":10,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/88\/revisions"}],"predecessor-version":[{"id":744,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/88\/revisions\/744"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=88"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}