{"id":1989,"date":"2022-05-03T05:57:19","date_gmt":"2022-05-03T05:57:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1989"},"modified":"2026-01-22T18:19:38","modified_gmt":"2026-01-22T18:19:38","slug":"4-4-algebraic-analysis-on-x-intercepts","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/4-4-algebraic-analysis-on-x-intercepts\/","title":{"raw":"4.4: Zeros of a Quadratic Function","rendered":"4.4: Zeros of a Quadratic Function"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Determine the zeros of a given quadratic function by factoring.<\/li>\r\n \t<li>Determine the zeros of a given quadratic function by completing the square<\/li>\r\n \t<li>Determine the zeros of a given quadratic function by using the quadratic formula<\/li>\r\n \t<li>Understand the relationship between zeros and [latex]x[\/latex]-intercepts<\/li>\r\n \t<li>Understand that quadratic functions with complex zeros have no [latex]x[\/latex]-intercepts<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring<\/h2>\r\nIn chapter 3, we learned several factoring methods and used them to find the zeros of polynomial functions. A quadratic function is a special case of a polynomial function. It is a trinomial of the form [latex]f(x)=ax^2+bx+c[\/latex], where [latex]a, b, c[\/latex] are real numbers and [latex]a\\ne 0[\/latex]. Consequently, we have already learned how to determine the zeros of a quadratic function by factoring.\r\n\r\nFor example, to determine the zeros of [latex]f(x)=9x^2-4[\/latex], we set [latex]f(x)=0[\/latex] and solve for [latex]x[\/latex] by factoring and using the zero-product property:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}9x^2-4 &amp;=0\\\\(3x-2)(3x+2)&amp;=0\\\\3x-2=0\\text{ or }3x+2&amp;=0\\\\x=\\dfrac{2}{3}\\text{ or }x&amp;=-\\dfrac{2}{3}\\end{aligned}[\/latex]<\/p>\r\nThe zeros of the function\u00a0[latex]f(x)=9x^2-4[\/latex] are [latex]x=-\\dfrac{2}{3},\\;\\dfrac{2}{3}[\/latex] or [latex]x=\\pm\\dfrac{2}{3}[\/latex], where the symbol [latex]\\pm[\/latex] is read \"plus or minus\" or \"positive or negative\".\r\n\r\nWe also learned that the zeros correspond with the [latex]x[\/latex]-coordinates of the [latex]x[\/latex]-intercepts on the graphs of the functions. So the [latex]x[\/latex]-intercepts of [latex]f(x)=9x^2-4[\/latex] are [latex]\\left(-\\dfrac{2}{3},\\;0\\right)[\/latex] and [latex]\\left(\\dfrac{2}{3}, 0\\right)[\/latex].\r\n\r\nWe will not repeat those factoring methods here. Factoring and using the zero-product property works well when the function factors. But in most cases, functions do not factor, so other methods need to be employed. In this section, we will focus on two methods that can be used to determine the zeros (and consequently the [latex]x[\/latex]-intercepts) of any quadratic function that may or may not factor.\r\n<h2>Completing the Square<\/h2>\r\nWe just learned in section 4.3 how to convert a quadratic function from its standard form [latex]f(x)=ax^2+bx+c[\/latex] into its vertex form [latex]f(x)=a(x-h)^2+k[\/latex] to determine the vertex of the function. In this section, we will take advantage of that new found knowledge to determine the zeros of a quadratic function.\r\n\r\nFor example, to determine the zeros of the function [latex]f(x)=x^2 - 4[\/latex] we set the function equal to 0. To solve the equation, we may isolate the perfect square, so we have a perfect square equal to a constant:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2 - 4 &amp;= 0\\\\x^2&amp;=4\\end{aligned}[\/latex]<\/p>\r\nTo solve this equation we first need to learn about the square root property.\r\n<h3>The Square Root Property<\/h3>\r\nThe square root property of equality tells us that if we have an equation [latex]A=B[\/latex], we can take the square root of each side of the equation to get an equivalent equation, [latex]\\sqrt{A}=\\sqrt{B}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The Square Root Property of Equality<\/h3>\r\n<p style=\"text-align: center;\">If [latex]A=B[\/latex], then\u00a0[latex]\\sqrt{A}=\\sqrt{B}[\/latex]<\/p>\r\n\r\n<\/div>\r\nSo, since we have [latex]x^2=4[\/latex], then [latex]\\sqrt{x^2}=\\sqrt{4}[\/latex].\r\n\r\nWe already know that [latex]\\sqrt{4}=2[\/latex], but what is [latex]\\sqrt{x^2}[\/latex]?\r\n\r\nThere are two cases to consider: [latex]x\u22650[\/latex] and [latex]x&lt;0[\/latex].\r\n\r\nSuppose [latex]x=9[\/latex], then [latex]x^2=(9)^2=81[\/latex]. Then [latex]\\sqrt{x^2}=\\sqrt{81}=9[\/latex]. And since [latex]9=x[\/latex], [latex]\\sqrt{x^2}=x[\/latex] when [latex]x=9[\/latex].\r\n\r\nIn fact, we could use any value of [latex]x\u22650[\/latex] and get the same result. So, [latex]\\sqrt{x^2}=x[\/latex] when [latex]x\u22650[\/latex].\r\n\r\nBut what happens when [latex]x&lt;0[\/latex]?\r\n\r\nSuppose [latex]x=-7[\/latex],\u00a0then [latex]x^2=(-7)^2=49[\/latex]. Then [latex]\\sqrt{x^2}=\\sqrt{49}=7[\/latex]. But [latex]x=-7[\/latex], not [latex]7[\/latex]! So when [latex]x&lt;0[\/latex], [latex]\\sqrt{x^2}=-x[\/latex], which is a positive answer when [latex]x[\/latex] is negative.\r\n\r\nPutting theses two results together, and not knowing whether [latex]x[\/latex] is positive or negative, we have, [latex]\\sqrt{x^2}=| x |[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>SQUARE ROOTS<\/h3>\r\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2}=| x |[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: left;\">Going back to our example,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2&amp;=4\\\\\\sqrt{x^2}&amp;=\\sqrt{4}\\\\|x|&amp;=2\\\\x&amp;=\\pm2\\end{aligned}[\/latex]<\/p>\r\nTherefore, the zeros are [latex]x=-2[\/latex] and [latex]x=2[\/latex]. These correspond with the two\u00a0[latex]x[\/latex]-intercepts on the graph of the function (2, 0) and (\u20132, 0).\r\n<p style=\"text-align: left;\">Notice that we have an absolute value equal to a constant, (i.e., [latex]|x|=2[\/latex]). This means that [latex]x[\/latex] will equal either positive or negative of the constant, (i.e., [latex]x=\\pm2[\/latex]). This is because the absolute value of a constant and its opposite are the same, (i.e., [latex]|2|=|-2|=2[\/latex].<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nDetermine the zeros of\u00a0[latex]g(x)=4(x-3)^2 - 32[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n<h4>Solution<\/h4>\r\nWe start by setting [latex]g(x)=0[\/latex], then isolating the perfect square:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}4(x-3)^2 - 32 &amp;= 0\\\\4(x-3)^2&amp;=32\\\\(x-3)^2&amp;=8\\end{aligned}[\/latex]<\/p>\r\nWe can now take the square root of both sides of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{(x-3)^2}&amp;=\\sqrt{8}\\\\|x-3|&amp;=\\sqrt{4\\cdot 2}\\\\x-3&amp;=\\pm2\\sqrt{2}\\\\x&amp;=3\\pm2\\sqrt{2}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe zeros of the function are [latex]x=3+2\\sqrt{2}[\/latex] and [latex]x=3-2\\sqrt{2}[\/latex].\r\n\r\nThe [latex]x[\/latex]-coordinates are therefore [latex]\\left(3+2\\sqrt{2},\\,0\\right)[\/latex] and\u00a0[latex]\\left(3-2\\sqrt{2},\\,0\\right)[\/latex].\r\n\r\n<\/div>\r\nNotice that when we simplified [latex]\\sqrt{8}[\/latex] in Example 1, we rewrote\u00a0[latex]\\sqrt{8}[\/latex] as\u00a0[latex]\\sqrt{4\\cdot 2}[\/latex]. We did this because to simplify square roots we look for perfect square factors in the radicand. Since [latex]4[\/latex] is the largest perfect square factor of [latex]8[\/latex], we write [latex]8=4\\cdot 2[\/latex]. We can then use the product property of radicals [latex]\\left (\\sqrt{ab}=\\sqrt{a}\\sqrt{b}\\right )[\/latex] to simplify:\u00a0[latex]\\sqrt{8}=\\sqrt{4\\cdot 2}=\\sqrt{4}\\sqrt{2}=2\\sqrt{2}[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=2(x+4)^2 - 50[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n<h4>Solution<\/h4>\r\nStart by setting [latex]f(x)=0[\/latex], the isolating the perfect square:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2(x+4)^2 - 50&amp;=0\\\\2(x+4)^2&amp;=50\\\\(x+4)^2&amp;=25\\end{aligned}[\/latex]<\/p>\r\nWe can now take the square root of both sides of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{(x+4)^2}&amp;=\\sqrt{25}\\\\|x+4|&amp;=5\\\\x+4&amp;=\\pm5\\\\x&amp;=-4\\pm5\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">This results in two equations:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=-4+5\\\\&amp;=1\\end{aligned}[\/latex]\u00a0 \u00a0and\u00a0 \u00a0[latex]\\begin{aligned}x&amp;=-4-5\\\\&amp;=-9\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe zeros of the function are\u00a0[latex]x=1[\/latex] and [latex]x=-9[\/latex].\r\n\r\nThe [latex]x[\/latex]-intercepts are therefore [latex](1, 0)[\/latex] and [latex](-9, 0)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=3(x-1)^2 - 27[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm308\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm308\"]\r\n\r\nZeros are [latex]x=-2[\/latex] and [latex]x=4[\/latex].\r\n\r\n[latex]x[\/latex]-intercepts are\u00a0(\u20132, 0) and (4, 0).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=2(x+3)^2 - 36[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm309\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm309\"]\r\n\r\nZeros are <span style=\"font-size: 1rem; text-align: initial;\">[latex]x=-3-3\\sqrt{2}[\/latex] and [latex]x=-3+3\\sqrt{2}[\/latex].<\/span>\r\n\r\n[latex]x[\/latex]-intercepts are [latex]\\left(-3-3\\sqrt{2}, 0\\right)[\/latex] and\u00a0[latex]\\left(-3+3\\sqrt{2}, 0\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf the function is given in standard form, we first need to convert it to\u00a0vertex form using completing the square.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=2x^2+12x-20[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n<h4>Solution<\/h4>\r\nStart by using completing the square to convert the function into vertex form:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=2x^2+12x-20\\\\&amp;=2\\left(x^2+6x\\right)-20\\\\&amp;=2\\left[(x+3)^2-9\\right]-20\\\\&amp;=2(x+3)^2-18-20\\\\&amp;=2(x+3)^2-38\\end{aligned}[\/latex]<\/p>\r\nNow we can set the function equal to zero:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2(x+3)^2-38&amp;=0\\\\2(x+3)^2&amp;=38\\\\(x+3)^2&amp;=19\\\\\\sqrt{(x+3)^2}&amp;=\\sqrt{19}\\\\|x+3|&amp;=\\sqrt{19}\\\\x+3&amp;=\\pm\\sqrt{19}\\\\x&amp;=-3\\pm\\sqrt{19}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe zeros of the function are\u00a0[latex]x=-3-\\sqrt{19}[\/latex] and [latex]x=-3+\\sqrt{19}[\/latex].\r\n\r\nThe [latex]x[\/latex]-intercepts are therefore [latex](-3-\\sqrt{19}, 0)[\/latex] and [latex](-3+\\sqrt{19}, 0)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=x^2+4x-12[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm983\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm983\"]\r\n\r\nThe zeros are [latex]x=-6[\/latex] and [latex]x=2[\/latex].\r\n\r\nThe [latex]x[\/latex]-intercepts are [latex](-6, 0)[\/latex] and [latex](2, 0)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nDetermine the zeros of\u00a0[latex]f(x)=3x^2+17x+10[\/latex], then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm984\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm984\"]\r\n\r\nThe zeros are [latex]x=-\\dfrac{2}{3}[\/latex] and [latex]x=-5[\/latex].\r\n\r\nThe [latex]x[\/latex]-intercepts are [latex](-\\dfrac{2}{3}, 0)[\/latex] and [latex](-5, 0)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Imaginary Numbers and Complex Zeros<\/h3>\r\nWhile trying to determine the zeros of a quadratic function, it is possible that we end up with an equation where the perfect square is equal to a negative number. For example, determining the zeros of the the function [latex]f(x)=x^2+4[\/latex] leads to the equation\u00a0<span style=\"text-align: center; font-size: 1em;\">[latex]x^2 = -4.[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0Does this equation have a solution?<\/span>\r\n\r\nIn the domain of real numbers, it is impossible to find a number whose square is a negative number. The square of every real number, whether positive or negative, is positive. Therefore, there are no real number solutions for this equation.\r\n\r\nHowever, the square root of a negative number is defined in the set of\u00a0<strong><em>imaginary numbers<\/em><\/strong>. The square root of \u20131 is defined as the letter [latex]i[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\sqrt{-1} = i[\/latex]<\/p>\r\nConsequently, all negative real numbers have a square root that is an imaginary number:\r\n<p style=\"text-align: center;\">[latex]\\sqrt{-3} = \\sqrt{-1}\\cdot\\sqrt{3} = i\\sqrt{3}[\/latex] (we always write the [latex]i[\/latex] in front of the radical)<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sqrt{-100} = \\sqrt{-1}\\cdot\\sqrt{100} = i\\sqrt{100} = 10i[\/latex]<\/p>\r\nWhen we combine through addition a real number [latex]a[\/latex] and an imaginary number [latex]bi[\/latex], where [latex]a,\\;b[\/latex] are real numbers, we form a\u00a0<em><strong>complex number<\/strong><\/em> [latex]a+bi[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>COMPLEX NUMBERS<\/h3>\r\nThe set of complex numbers [latex]\\mathbb{C}[\/latex] is the union of the set of real numbers [latex]\\mathbb{R}[\/latex] and the set of imaginary numbers [latex]\\{bi\\;|\\;i=\\sqrt{-1}\\;,\\;b\\in \\mathbb{R}\\;\\}[\/latex].\r\n\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex], [latex]a+bi[\/latex] is a complex number.\r\n\r\n<\/div>\r\nComplex zeros will show up when we end up with a perfect square equal to a negative number.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nDetermine the zeros of the function [latex]f(x)=\\left(x+3\\right)^2+16[\/latex].\r\n<h4>Solution<\/h4>\r\nTo determine the zeros we set the function equal to zero and solve the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x+3)^2+16&amp;=0\\\\(x+3)^2&amp;=-16\\\\\\sqrt{(x+3)^2}&amp;=\\sqrt{-16}\\\\|x+3|&amp;=4i\\\\x+3&amp;=\\pm4i\\\\x&amp;=-3\\pm4i\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe result is two complex zeros [latex]x=-3-4i[\/latex] and\u00a0[latex]x=-3+4i[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=x^2-4x+24[\/latex].\r\n<h4>Solution<\/h4>\r\nWe first convert the function to vertex form by completing the square:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2-4x+24&amp;=(x-2)^2-4+24\\\\&amp;=(x-2)^2+20\\end{aligned}[\/latex]<\/p>\r\nTo determine the zeros we set the function equal to zero and solve the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x-2)^2+20&amp;=0\\\\(x-2)^2&amp;=-20\\\\\\sqrt{(x-2)^2}&amp;=\\sqrt{-20}\\\\|x-2|&amp;=\\sqrt{-4}\\sqrt{5}\\\\x-2&amp;=\\pm2i\\sqrt{5}\\\\x&amp;=2\\pm2i\\sqrt{5}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe result is two complex zeros [latex]x=2-2i\\sqrt{5}[\/latex] and\u00a0[latex]x=2+2i\\sqrt{5}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=x^2-6x+30[\/latex].\r\n\r\n[reveal-answer q=\"hjm171\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm171\"]\r\n\r\nThe zeros are [latex]x=3-i\\sqrt{21}[\/latex] and [latex]x=3+i\\sqrt{21}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nA function having complex zeros is not unusual, but what does this mean for the [latex]x[\/latex]-intercepts of the graph of such a function?\r\n\r\nLet's look at the graphs of the functions from examples 4 and 5.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%;\" colspan=\"2\">Graphs of parabolas with complex zeros<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\"><img class=\"aligncenter wp-image-2721 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-300x300.png\" alt=\"Parabola opening upward with vertex at (-3,16)\" width=\"300\" height=\"300\" \/>\r\n<p style=\"text-align: center;\">[latex]g(x)=x^2+6x+25[\/latex] with<\/p>\r\n<p style=\"text-align: center;\">complex zeros [latex]x=-3-4i[\/latex] and\u00a0[latex]x=-3+4i[\/latex]<\/p>\r\n<\/td>\r\n<td style=\"width: 50%;\">&nbsp;\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-2723 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-300x300.png\" alt=\"parabola opening upward with vertex at (2,20)\" width=\"300\" height=\"300\" \/>[latex]g(x)=x^2-4x+24[\/latex] with<\/p>\r\n<p style=\"text-align: center;\">complex zeros [latex]x=2-2i\\sqrt{5}[\/latex] and\u00a0[latex]x=2+2i\\sqrt{5}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\" colspan=\"2\">Figure 1. Graphs of functions with complex zeros.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere are no [latex]x[\/latex]-intercepts when a quadratic function has complex zeros. This is because the coordinate system, and in particular the [latex]x[\/latex]-axis, graphs only real numbers.\r\n<div class=\"textbox shaded\">\r\n<h3>graphs of quadratic functions with complex zeros<\/h3>\r\nThe graph of a quadratic function with complex zeros has no [latex]x[\/latex]-intercepts; the parabola never intersects the [latex]x[\/latex]-axis.\r\n\r\n<\/div>\r\n<h2>The Quadratic Formula<\/h2>\r\nAnother way to determine the zeros of a quadratic function\u00a0<span style=\"font-size: 1em;\">[latex]f(x) = ax^2 + bx + c[\/latex]<\/span><span style=\"font-size: 1em;\">\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is to use the quadratic formula.<\/span>\r\n<div class=\"textbox shaded\">\r\n<h3>The quadratic formula<\/h3>\r\n<p style=\"text-align: center;\">For any quadratic function [latex]f(x)=ax^2+bx+c[\/latex], the zeros of the function are given by<\/p>\r\n<p style=\"text-align: center;\">[latex]x = \\dfrac{-b \\pm\\sqrt{b^2-4ac}}{2a}[\/latex]<\/p>\r\n\r\n<\/div>\r\nTo use the quadratic equation, we need a quadratic function in the standard form [latex]f(x)=ax^2+bx+c[\/latex] so that we can pick out the values of [latex]a,\\;b,\\;c[\/latex].\r\n\r\nFor example if we are asked to determine the zeros of the function\u00a0[latex]h(x)=3x^2-5x+1[\/latex], we evaluate the quadratic equation when [latex]a = 3, b = -5, c = 1[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x &amp;=\\dfrac{-(-5) \\pm\\sqrt{(-5)^2-4(3)(1)}}{2(3)} \\\\&amp;= \\dfrac{5\\pm\\sqrt{25-12}}{6} \\\\=&amp; \\dfrac{5\\pm\\sqrt{13}}{6} \\end{aligned}[\/latex]<\/p>\r\nTherefore, the zeros are [latex]x=\\dfrac{5 - \\sqrt{13}}{6}[\/latex] and\u00a0[latex]x=\\dfrac{5 + \\sqrt{13}}{6}[\/latex].\r\n\r\nIf we were asked to find the\u00a0[latex]x[\/latex]-intercepts, they are [latex](\\dfrac{5 + \\sqrt{13}}{6}, 0)[\/latex] and\u00a0[latex](\\dfrac{5 - \\sqrt{13}}{6}, 0)[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=x^2-4x+2[\/latex]. Then state the [latex]x[\/latex]-intercepts.\r\n<h4>Solution<\/h4>\r\nTo use the quadratic formula we need [latex]a=1,\\;b=-4,\\;c=2[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&amp;=\\dfrac{-(-4)\\pm\\sqrt{(-4)^2-4(1)(2)}}{2(1)}\\\\&amp;=\\dfrac{4\\pm\\sqrt{16-8}}{2}\\\\&amp;=\\dfrac{4\\pm\\sqrt{8}}{2}\\\\&amp;=\\dfrac{4\\pm2\\sqrt{2}}{2}\\\\&amp;=2\\pm\\sqrt{2}\\\\&amp;=2+\\sqrt{2},\\;\\;2-\\sqrt{2}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe zeros of the function are [latex]x=2+\\sqrt{2}[\/latex] and [latex]x=2-\\sqrt{2}[\/latex].\r\n\r\nThe [latex]x[\/latex]-intercepts are [latex](2+\\sqrt{2}, 0)[\/latex] and [latex](2-\\sqrt{2}, 0)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 7<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=2x^2-x+5[\/latex]. Then state the [latex]x[\/latex]-intercepts.\r\n<h4>Solution<\/h4>\r\nFirst pick out [latex]a=2,\\;b=-1,\\;c=5[\/latex], then evaluate the quadratic formula:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&amp;=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&amp;=\\dfrac{-(-1)\\pm\\sqrt{(-1)^2-4(2)(5)}}{2(2)}\\\\&amp;=\\dfrac{1\\pm\\sqrt{1-40}}{4}\\\\&amp;=\\dfrac{1\\pm\\sqrt{-39}}{4}\\\\&amp;=\\dfrac{1\\pm i\\sqrt{39}}{4}\\\\&amp;=\\dfrac{1}{4}\\pm\\dfrac{i\\sqrt{39}}{4}\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe zeros of the function are [latex]x=\\dfrac{1}{4}-\\dfrac{i\\sqrt{39}}{4}[\/latex] and [latex]\\dfrac{1}{4}+\\dfrac{i\\sqrt{39}}{4}[\/latex].\r\n\r\nThere are no [latex]x[\/latex]-intercepts since the zeros are complex.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=x^2-3x+2[\/latex]. Then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm966\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm966\"]\r\n\r\nZeros are [latex]x=1[\/latex] and [latex]x=2[\/latex].\r\n\r\n[latex]x[\/latex]-intercepts are\u00a0[latex](1, 0)[\/latex] and [latex](2, 0)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 7<\/h3>\r\nDetermine the zeros of the function [latex]g(x)=3x^2+12[\/latex]. Then state the [latex]x[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"hjm720\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm720\"]\r\n\r\nZeros are [latex]x=-2i[\/latex] and\u00a0[latex]x=2i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Proof of the Quadratic Formula<\/h3>\r\nThis quadratic formula may be derived by converting the function [latex]f(x) = ax^2 + bx + c[\/latex] into vertex form by completing the square, and then setting [latex]f(x)=0[\/latex] and solving for\u00a0[latex]x[\/latex]. The following proof shows how this formula is derived.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=ax^2+bx+c\\\\&amp;=a\\left[x^2 + \\frac{b}{a}x\\right] + c\\\\&amp;=a\\left[\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2 - \\left(\\dfrac{b}{2a}\\right)^2\\right] + c\\\\&amp;=a\\left(x+\\left(\\frac{b}{2a}\\right)\\right)^2 - a\\left(\\frac{b}{2a}\\right)^2 + c\\\\&amp;=a\\left(x+\\left(\\frac{b}{2a}\\right)\\right)^2 - \\frac{b^2}{4a} + c\\end{aligned}[\/latex]<\/p>\r\nTo find the zeros, set the function equal to zero, and then solve for\u00a0[latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}a\\left(x+\\left(d\\frac{b}{2a}\\right)\\right)^2 - \\dfrac{b^2}{4a} + c &amp;= 0\\\\a\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2&amp;=\\dfrac{b^2}{4a} - c\\\\a\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2 &amp;=\\dfrac{b^2-4ac}{4a}\\\\x+\\left(\\dfrac{b}{2a}\\right)^2 &amp;= \\dfrac{b^2-4ac}{4a^2}\\\\\\sqrt{\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2} &amp;=\\sqrt{\\dfrac{b^2-4ac}{4a^2}}\\\\\\mathopen| x+\\dfrac{b}{2a}\\mathclose| &amp;=\\dfrac{\\sqrt{b^2-4ac}}{\\sqrt{4a^2}}\\\\x+\\dfrac{b}{2a}&amp;=\\pm\\dfrac{\\sqrt{b^2-4ac}}{2|a|}\\\\x &amp;= -\\frac{b}{2a}\\pm\\frac{\\sqrt{b^2-4ac}}{\\pm2a}\\\\x &amp;= -\\dfrac{b}{2a} \\pm \\dfrac{\\sqrt{b^2-4ac}}{2a}\\\\x &amp;= \\dfrac{-b \\pm\\sqrt{b^2-4ac}}{2a}\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: right;\"><a href=\"https:\/\/www.merriam-webster.com\/dictionary\/quod%20erat%20demonstrandum\">Q.E.D.<\/a><\/p>\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Determine the zeros of a given quadratic function by factoring.<\/li>\n<li>Determine the zeros of a given quadratic function by completing the square<\/li>\n<li>Determine the zeros of a given quadratic function by using the quadratic formula<\/li>\n<li>Understand the relationship between zeros and [latex]x[\/latex]-intercepts<\/li>\n<li>Understand that quadratic functions with complex zeros have no [latex]x[\/latex]-intercepts<\/li>\n<\/ul>\n<\/div>\n<h2>Factoring<\/h2>\n<p>In chapter 3, we learned several factoring methods and used them to find the zeros of polynomial functions. A quadratic function is a special case of a polynomial function. It is a trinomial of the form [latex]f(x)=ax^2+bx+c[\/latex], where [latex]a, b, c[\/latex] are real numbers and [latex]a\\ne 0[\/latex]. Consequently, we have already learned how to determine the zeros of a quadratic function by factoring.<\/p>\n<p>For example, to determine the zeros of [latex]f(x)=9x^2-4[\/latex], we set [latex]f(x)=0[\/latex] and solve for [latex]x[\/latex] by factoring and using the zero-product property:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}9x^2-4 &=0\\\\(3x-2)(3x+2)&=0\\\\3x-2=0\\text{ or }3x+2&=0\\\\x=\\dfrac{2}{3}\\text{ or }x&=-\\dfrac{2}{3}\\end{aligned}[\/latex]<\/p>\n<p>The zeros of the function\u00a0[latex]f(x)=9x^2-4[\/latex] are [latex]x=-\\dfrac{2}{3},\\;\\dfrac{2}{3}[\/latex] or [latex]x=\\pm\\dfrac{2}{3}[\/latex], where the symbol [latex]\\pm[\/latex] is read &#8220;plus or minus&#8221; or &#8220;positive or negative&#8221;.<\/p>\n<p>We also learned that the zeros correspond with the [latex]x[\/latex]-coordinates of the [latex]x[\/latex]-intercepts on the graphs of the functions. So the [latex]x[\/latex]-intercepts of [latex]f(x)=9x^2-4[\/latex] are [latex]\\left(-\\dfrac{2}{3},\\;0\\right)[\/latex] and [latex]\\left(\\dfrac{2}{3}, 0\\right)[\/latex].<\/p>\n<p>We will not repeat those factoring methods here. Factoring and using the zero-product property works well when the function factors. But in most cases, functions do not factor, so other methods need to be employed. In this section, we will focus on two methods that can be used to determine the zeros (and consequently the [latex]x[\/latex]-intercepts) of any quadratic function that may or may not factor.<\/p>\n<h2>Completing the Square<\/h2>\n<p>We just learned in section 4.3 how to convert a quadratic function from its standard form [latex]f(x)=ax^2+bx+c[\/latex] into its vertex form [latex]f(x)=a(x-h)^2+k[\/latex] to determine the vertex of the function. In this section, we will take advantage of that new found knowledge to determine the zeros of a quadratic function.<\/p>\n<p>For example, to determine the zeros of the function [latex]f(x)=x^2 - 4[\/latex] we set the function equal to 0. To solve the equation, we may isolate the perfect square, so we have a perfect square equal to a constant:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2 - 4 &= 0\\\\x^2&=4\\end{aligned}[\/latex]<\/p>\n<p>To solve this equation we first need to learn about the square root property.<\/p>\n<h3>The Square Root Property<\/h3>\n<p>The square root property of equality tells us that if we have an equation [latex]A=B[\/latex], we can take the square root of each side of the equation to get an equivalent equation, [latex]\\sqrt{A}=\\sqrt{B}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The Square Root Property of Equality<\/h3>\n<p style=\"text-align: center;\">If [latex]A=B[\/latex], then\u00a0[latex]\\sqrt{A}=\\sqrt{B}[\/latex]<\/p>\n<\/div>\n<p>So, since we have [latex]x^2=4[\/latex], then [latex]\\sqrt{x^2}=\\sqrt{4}[\/latex].<\/p>\n<p>We already know that [latex]\\sqrt{4}=2[\/latex], but what is [latex]\\sqrt{x^2}[\/latex]?<\/p>\n<p>There are two cases to consider: [latex]x\u22650[\/latex] and [latex]x<0[\/latex].\n\nSuppose [latex]x=9[\/latex], then [latex]x^2=(9)^2=81[\/latex]. Then [latex]\\sqrt{x^2}=\\sqrt{81}=9[\/latex]. And since [latex]9=x[\/latex], [latex]\\sqrt{x^2}=x[\/latex] when [latex]x=9[\/latex].\n\nIn fact, we could use any value of [latex]x\u22650[\/latex] and get the same result. So, [latex]\\sqrt{x^2}=x[\/latex] when [latex]x\u22650[\/latex].\n\nBut what happens when [latex]x<0[\/latex]?\n\nSuppose [latex]x=-7[\/latex],\u00a0then [latex]x^2=(-7)^2=49[\/latex]. Then [latex]\\sqrt{x^2}=\\sqrt{49}=7[\/latex]. But [latex]x=-7[\/latex], not [latex]7[\/latex]! So when [latex]x<0[\/latex], [latex]\\sqrt{x^2}=-x[\/latex], which is a positive answer when [latex]x[\/latex] is negative.\n\nPutting theses two results together, and not knowing whether [latex]x[\/latex] is positive or negative, we have, [latex]\\sqrt{x^2}=| x |[\/latex].\n\n\n<div class=\"textbox shaded\">\n<h3>SQUARE ROOTS<\/h3>\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2}=| x |[\/latex]<\/p>\n<\/div>\n<p style=\"text-align: left;\">Going back to our example,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2&=4\\\\\\sqrt{x^2}&=\\sqrt{4}\\\\|x|&=2\\\\x&=\\pm2\\end{aligned}[\/latex]<\/p>\n<p>Therefore, the zeros are [latex]x=-2[\/latex] and [latex]x=2[\/latex]. These correspond with the two\u00a0[latex]x[\/latex]-intercepts on the graph of the function (2, 0) and (\u20132, 0).<\/p>\n<p style=\"text-align: left;\">Notice that we have an absolute value equal to a constant, (i.e., [latex]|x|=2[\/latex]). This means that [latex]x[\/latex] will equal either positive or negative of the constant, (i.e., [latex]x=\\pm2[\/latex]). This is because the absolute value of a constant and its opposite are the same, (i.e., [latex]|2|=|-2|=2[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Determine the zeros of\u00a0[latex]g(x)=4(x-3)^2 - 32[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<h4>Solution<\/h4>\n<p>We start by setting [latex]g(x)=0[\/latex], then isolating the perfect square:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}4(x-3)^2 - 32 &= 0\\\\4(x-3)^2&=32\\\\(x-3)^2&=8\\end{aligned}[\/latex]<\/p>\n<p>We can now take the square root of both sides of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{(x-3)^2}&=\\sqrt{8}\\\\|x-3|&=\\sqrt{4\\cdot 2}\\\\x-3&=\\pm2\\sqrt{2}\\\\x&=3\\pm2\\sqrt{2}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The zeros of the function are [latex]x=3+2\\sqrt{2}[\/latex] and [latex]x=3-2\\sqrt{2}[\/latex].<\/p>\n<p>The [latex]x[\/latex]-coordinates are therefore [latex]\\left(3+2\\sqrt{2},\\,0\\right)[\/latex] and\u00a0[latex]\\left(3-2\\sqrt{2},\\,0\\right)[\/latex].<\/p>\n<\/div>\n<p>Notice that when we simplified [latex]\\sqrt{8}[\/latex] in Example 1, we rewrote\u00a0[latex]\\sqrt{8}[\/latex] as\u00a0[latex]\\sqrt{4\\cdot 2}[\/latex]. We did this because to simplify square roots we look for perfect square factors in the radicand. Since [latex]4[\/latex] is the largest perfect square factor of [latex]8[\/latex], we write [latex]8=4\\cdot 2[\/latex]. We can then use the product property of radicals [latex]\\left (\\sqrt{ab}=\\sqrt{a}\\sqrt{b}\\right )[\/latex] to simplify:\u00a0[latex]\\sqrt{8}=\\sqrt{4\\cdot 2}=\\sqrt{4}\\sqrt{2}=2\\sqrt{2}[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=2(x+4)^2 - 50[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<h4>Solution<\/h4>\n<p>Start by setting [latex]f(x)=0[\/latex], the isolating the perfect square:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2(x+4)^2 - 50&=0\\\\2(x+4)^2&=50\\\\(x+4)^2&=25\\end{aligned}[\/latex]<\/p>\n<p>We can now take the square root of both sides of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\sqrt{(x+4)^2}&=\\sqrt{25}\\\\|x+4|&=5\\\\x+4&=\\pm5\\\\x&=-4\\pm5\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">This results in two equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=-4+5\\\\&=1\\end{aligned}[\/latex]\u00a0 \u00a0and\u00a0 \u00a0[latex]\\begin{aligned}x&=-4-5\\\\&=-9\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The zeros of the function are\u00a0[latex]x=1[\/latex] and [latex]x=-9[\/latex].<\/p>\n<p>The [latex]x[\/latex]-intercepts are therefore [latex](1, 0)[\/latex] and [latex](-9, 0)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=3(x-1)^2 - 27[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm308\">Show Answer<\/span><\/p>\n<div id=\"qhjm308\" class=\"hidden-answer\" style=\"display: none\">\n<p>Zeros are [latex]x=-2[\/latex] and [latex]x=4[\/latex].<\/p>\n<p>[latex]x[\/latex]-intercepts are\u00a0(\u20132, 0) and (4, 0).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=2(x+3)^2 - 36[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm309\">Show Answer<\/span><\/p>\n<div id=\"qhjm309\" class=\"hidden-answer\" style=\"display: none\">\n<p>Zeros are <span style=\"font-size: 1rem; text-align: initial;\">[latex]x=-3-3\\sqrt{2}[\/latex] and [latex]x=-3+3\\sqrt{2}[\/latex].<\/span><\/p>\n<p>[latex]x[\/latex]-intercepts are [latex]\\left(-3-3\\sqrt{2}, 0\\right)[\/latex] and\u00a0[latex]\\left(-3+3\\sqrt{2}, 0\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>If the function is given in standard form, we first need to convert it to\u00a0vertex form using completing the square.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=2x^2+12x-20[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<h4>Solution<\/h4>\n<p>Start by using completing the square to convert the function into vertex form:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=2x^2+12x-20\\\\&=2\\left(x^2+6x\\right)-20\\\\&=2\\left[(x+3)^2-9\\right]-20\\\\&=2(x+3)^2-18-20\\\\&=2(x+3)^2-38\\end{aligned}[\/latex]<\/p>\n<p>Now we can set the function equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2(x+3)^2-38&=0\\\\2(x+3)^2&=38\\\\(x+3)^2&=19\\\\\\sqrt{(x+3)^2}&=\\sqrt{19}\\\\|x+3|&=\\sqrt{19}\\\\x+3&=\\pm\\sqrt{19}\\\\x&=-3\\pm\\sqrt{19}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The zeros of the function are\u00a0[latex]x=-3-\\sqrt{19}[\/latex] and [latex]x=-3+\\sqrt{19}[\/latex].<\/p>\n<p>The [latex]x[\/latex]-intercepts are therefore [latex](-3-\\sqrt{19}, 0)[\/latex] and [latex](-3+\\sqrt{19}, 0)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=x^2+4x-12[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm983\">Show Answer<\/span><\/p>\n<div id=\"qhjm983\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are [latex]x=-6[\/latex] and [latex]x=2[\/latex].<\/p>\n<p>The [latex]x[\/latex]-intercepts are [latex](-6, 0)[\/latex] and [latex](2, 0)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Determine the zeros of\u00a0[latex]f(x)=3x^2+17x+10[\/latex], then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm984\">Show Answer<\/span><\/p>\n<div id=\"qhjm984\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are [latex]x=-\\dfrac{2}{3}[\/latex] and [latex]x=-5[\/latex].<\/p>\n<p>The [latex]x[\/latex]-intercepts are [latex](-\\dfrac{2}{3}, 0)[\/latex] and [latex](-5, 0)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Imaginary Numbers and Complex Zeros<\/h3>\n<p>While trying to determine the zeros of a quadratic function, it is possible that we end up with an equation where the perfect square is equal to a negative number. For example, determining the zeros of the the function [latex]f(x)=x^2+4[\/latex] leads to the equation\u00a0<span style=\"text-align: center; font-size: 1em;\">[latex]x^2 = -4.[\/latex]<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0Does this equation have a solution?<\/span><\/p>\n<p>In the domain of real numbers, it is impossible to find a number whose square is a negative number. The square of every real number, whether positive or negative, is positive. Therefore, there are no real number solutions for this equation.<\/p>\n<p>However, the square root of a negative number is defined in the set of\u00a0<strong><em>imaginary numbers<\/em><\/strong>. The square root of \u20131 is defined as the letter [latex]i[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-1} = i[\/latex]<\/p>\n<p>Consequently, all negative real numbers have a square root that is an imaginary number:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-3} = \\sqrt{-1}\\cdot\\sqrt{3} = i\\sqrt{3}[\/latex] (we always write the [latex]i[\/latex] in front of the radical)<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-100} = \\sqrt{-1}\\cdot\\sqrt{100} = i\\sqrt{100} = 10i[\/latex]<\/p>\n<p>When we combine through addition a real number [latex]a[\/latex] and an imaginary number [latex]bi[\/latex], where [latex]a,\\;b[\/latex] are real numbers, we form a\u00a0<em><strong>complex number<\/strong><\/em> [latex]a+bi[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>COMPLEX NUMBERS<\/h3>\n<p>The set of complex numbers [latex]\\mathbb{C}[\/latex] is the union of the set of real numbers [latex]\\mathbb{R}[\/latex] and the set of imaginary numbers [latex]\\{bi\\;|\\;i=\\sqrt{-1}\\;,\\;b\\in \\mathbb{R}\\;\\}[\/latex].<\/p>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex], [latex]a+bi[\/latex] is a complex number.<\/p>\n<\/div>\n<p>Complex zeros will show up when we end up with a perfect square equal to a negative number.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Determine the zeros of the function [latex]f(x)=\\left(x+3\\right)^2+16[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>To determine the zeros we set the function equal to zero and solve the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x+3)^2+16&=0\\\\(x+3)^2&=-16\\\\\\sqrt{(x+3)^2}&=\\sqrt{-16}\\\\|x+3|&=4i\\\\x+3&=\\pm4i\\\\x&=-3\\pm4i\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The result is two complex zeros [latex]x=-3-4i[\/latex] and\u00a0[latex]x=-3+4i[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=x^2-4x+24[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We first convert the function to vertex form by completing the square:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x^2-4x+24&=(x-2)^2-4+24\\\\&=(x-2)^2+20\\end{aligned}[\/latex]<\/p>\n<p>To determine the zeros we set the function equal to zero and solve the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x-2)^2+20&=0\\\\(x-2)^2&=-20\\\\\\sqrt{(x-2)^2}&=\\sqrt{-20}\\\\|x-2|&=\\sqrt{-4}\\sqrt{5}\\\\x-2&=\\pm2i\\sqrt{5}\\\\x&=2\\pm2i\\sqrt{5}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The result is two complex zeros [latex]x=2-2i\\sqrt{5}[\/latex] and\u00a0[latex]x=2+2i\\sqrt{5}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=x^2-6x+30[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm171\">Show Answer<\/span><\/p>\n<div id=\"qhjm171\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are [latex]x=3-i\\sqrt{21}[\/latex] and [latex]x=3+i\\sqrt{21}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>A function having complex zeros is not unusual, but what does this mean for the [latex]x[\/latex]-intercepts of the graph of such a function?<\/p>\n<p>Let&#8217;s look at the graphs of the functions from examples 4 and 5.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 50%;\" colspan=\"2\">Graphs of parabolas with complex zeros<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2721 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-300x300.png\" alt=\"Parabola opening upward with vertex at (-3,16)\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153119.107.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p style=\"text-align: center;\">[latex]g(x)=x^2+6x+25[\/latex] with<\/p>\n<p style=\"text-align: center;\">complex zeros [latex]x=-3-4i[\/latex] and\u00a0[latex]x=-3+4i[\/latex]<\/p>\n<\/td>\n<td style=\"width: 50%;\">&nbsp;<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2723 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-300x300.png\" alt=\"parabola opening upward with vertex at (2,20)\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-16T153417.645.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/>[latex]g(x)=x^2-4x+24[\/latex] with<\/p>\n<p style=\"text-align: center;\">complex zeros [latex]x=2-2i\\sqrt{5}[\/latex] and\u00a0[latex]x=2+2i\\sqrt{5}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\" colspan=\"2\">Figure 1. Graphs of functions with complex zeros.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There are no [latex]x[\/latex]-intercepts when a quadratic function has complex zeros. This is because the coordinate system, and in particular the [latex]x[\/latex]-axis, graphs only real numbers.<\/p>\n<div class=\"textbox shaded\">\n<h3>graphs of quadratic functions with complex zeros<\/h3>\n<p>The graph of a quadratic function with complex zeros has no [latex]x[\/latex]-intercepts; the parabola never intersects the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<h2>The Quadratic Formula<\/h2>\n<p>Another way to determine the zeros of a quadratic function\u00a0<span style=\"font-size: 1em;\">[latex]f(x) = ax^2 + bx + c[\/latex]<\/span><span style=\"font-size: 1em;\">\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is to use the quadratic formula.<\/span><\/p>\n<div class=\"textbox shaded\">\n<h3>The quadratic formula<\/h3>\n<p style=\"text-align: center;\">For any quadratic function [latex]f(x)=ax^2+bx+c[\/latex], the zeros of the function are given by<\/p>\n<p style=\"text-align: center;\">[latex]x = \\dfrac{-b \\pm\\sqrt{b^2-4ac}}{2a}[\/latex]<\/p>\n<\/div>\n<p>To use the quadratic equation, we need a quadratic function in the standard form [latex]f(x)=ax^2+bx+c[\/latex] so that we can pick out the values of [latex]a,\\;b,\\;c[\/latex].<\/p>\n<p>For example if we are asked to determine the zeros of the function\u00a0[latex]h(x)=3x^2-5x+1[\/latex], we evaluate the quadratic equation when [latex]a = 3, b = -5, c = 1[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x &=\\dfrac{-(-5) \\pm\\sqrt{(-5)^2-4(3)(1)}}{2(3)} \\\\&= \\dfrac{5\\pm\\sqrt{25-12}}{6} \\\\=& \\dfrac{5\\pm\\sqrt{13}}{6} \\end{aligned}[\/latex]<\/p>\n<p>Therefore, the zeros are [latex]x=\\dfrac{5 - \\sqrt{13}}{6}[\/latex] and\u00a0[latex]x=\\dfrac{5 + \\sqrt{13}}{6}[\/latex].<\/p>\n<p>If we were asked to find the\u00a0[latex]x[\/latex]-intercepts, they are [latex](\\dfrac{5 + \\sqrt{13}}{6}, 0)[\/latex] and\u00a0[latex](\\dfrac{5 - \\sqrt{13}}{6}, 0)[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=x^2-4x+2[\/latex]. Then state the [latex]x[\/latex]-intercepts.<\/p>\n<h4>Solution<\/h4>\n<p>To use the quadratic formula we need [latex]a=1,\\;b=-4,\\;c=2[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&=\\dfrac{-(-4)\\pm\\sqrt{(-4)^2-4(1)(2)}}{2(1)}\\\\&=\\dfrac{4\\pm\\sqrt{16-8}}{2}\\\\&=\\dfrac{4\\pm\\sqrt{8}}{2}\\\\&=\\dfrac{4\\pm2\\sqrt{2}}{2}\\\\&=2\\pm\\sqrt{2}\\\\&=2+\\sqrt{2},\\;\\;2-\\sqrt{2}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The zeros of the function are [latex]x=2+\\sqrt{2}[\/latex] and [latex]x=2-\\sqrt{2}[\/latex].<\/p>\n<p>The [latex]x[\/latex]-intercepts are [latex](2+\\sqrt{2}, 0)[\/latex] and [latex](2-\\sqrt{2}, 0)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 7<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=2x^2-x+5[\/latex]. Then state the [latex]x[\/latex]-intercepts.<\/p>\n<h4>Solution<\/h4>\n<p>First pick out [latex]a=2,\\;b=-1,\\;c=5[\/latex], then evaluate the quadratic formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x&=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&=\\dfrac{-(-1)\\pm\\sqrt{(-1)^2-4(2)(5)}}{2(2)}\\\\&=\\dfrac{1\\pm\\sqrt{1-40}}{4}\\\\&=\\dfrac{1\\pm\\sqrt{-39}}{4}\\\\&=\\dfrac{1\\pm i\\sqrt{39}}{4}\\\\&=\\dfrac{1}{4}\\pm\\dfrac{i\\sqrt{39}}{4}\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The zeros of the function are [latex]x=\\dfrac{1}{4}-\\dfrac{i\\sqrt{39}}{4}[\/latex] and [latex]\\dfrac{1}{4}+\\dfrac{i\\sqrt{39}}{4}[\/latex].<\/p>\n<p>There are no [latex]x[\/latex]-intercepts since the zeros are complex.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=x^2-3x+2[\/latex]. Then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm966\">Show Answer<\/span><\/p>\n<div id=\"qhjm966\" class=\"hidden-answer\" style=\"display: none\">\n<p>Zeros are [latex]x=1[\/latex] and [latex]x=2[\/latex].<\/p>\n<p>[latex]x[\/latex]-intercepts are\u00a0[latex](1, 0)[\/latex] and [latex](2, 0)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 7<\/h3>\n<p>Determine the zeros of the function [latex]g(x)=3x^2+12[\/latex]. Then state the [latex]x[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm720\">Show Answer<\/span><\/p>\n<div id=\"qhjm720\" class=\"hidden-answer\" style=\"display: none\">\n<p>Zeros are [latex]x=-2i[\/latex] and\u00a0[latex]x=2i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Proof of the Quadratic Formula<\/h3>\n<p>This quadratic formula may be derived by converting the function [latex]f(x) = ax^2 + bx + c[\/latex] into vertex form by completing the square, and then setting [latex]f(x)=0[\/latex] and solving for\u00a0[latex]x[\/latex]. The following proof shows how this formula is derived.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=ax^2+bx+c\\\\&=a\\left[x^2 + \\frac{b}{a}x\\right] + c\\\\&=a\\left[\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2 - \\left(\\dfrac{b}{2a}\\right)^2\\right] + c\\\\&=a\\left(x+\\left(\\frac{b}{2a}\\right)\\right)^2 - a\\left(\\frac{b}{2a}\\right)^2 + c\\\\&=a\\left(x+\\left(\\frac{b}{2a}\\right)\\right)^2 - \\frac{b^2}{4a} + c\\end{aligned}[\/latex]<\/p>\n<p>To find the zeros, set the function equal to zero, and then solve for\u00a0[latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}a\\left(x+\\left(d\\frac{b}{2a}\\right)\\right)^2 - \\dfrac{b^2}{4a} + c &= 0\\\\a\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2&=\\dfrac{b^2}{4a} - c\\\\a\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2 &=\\dfrac{b^2-4ac}{4a}\\\\x+\\left(\\dfrac{b}{2a}\\right)^2 &= \\dfrac{b^2-4ac}{4a^2}\\\\\\sqrt{\\left(x+\\left(\\dfrac{b}{2a}\\right)\\right)^2} &=\\sqrt{\\dfrac{b^2-4ac}{4a^2}}\\\\\\mathopen| x+\\dfrac{b}{2a}\\mathclose| &=\\dfrac{\\sqrt{b^2-4ac}}{\\sqrt{4a^2}}\\\\x+\\dfrac{b}{2a}&=\\pm\\dfrac{\\sqrt{b^2-4ac}}{2|a|}\\\\x &= -\\frac{b}{2a}\\pm\\frac{\\sqrt{b^2-4ac}}{\\pm2a}\\\\x &= -\\dfrac{b}{2a} \\pm \\dfrac{\\sqrt{b^2-4ac}}{2a}\\\\x &= \\dfrac{-b \\pm\\sqrt{b^2-4ac}}{2a}\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: right;\"><a href=\"https:\/\/www.merriam-webster.com\/dictionary\/quod%20erat%20demonstrandum\">Q.E.D.<\/a><\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1989\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Zeros of a Quadratic Function. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All graphs created using desmos graphing calculator. <strong>Authored by<\/strong>: Hazel McKenna . <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All examples. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All Try Its. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Zeros of a Quadratic Function\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created using desmos graphing calculator\",\"author\":\"Hazel McKenna \",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All examples\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All Try Its\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1989","chapter","type-chapter","status-publish","hentry"],"part":1691,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1989","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":59,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1989\/revisions"}],"predecessor-version":[{"id":4852,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1989\/revisions\/4852"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/1691"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1989\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=1989"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1989"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=1989"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=1989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}