{"id":204,"date":"2023-11-08T16:10:24","date_gmt":"2023-11-08T16:10:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-or-watch-the-quadratic-formula\/"},"modified":"2024-11-04T22:27:52","modified_gmt":"2024-11-04T22:27:52","slug":"6-2-the-quadratic-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-2-the-quadratic-formula\/","title":{"raw":"6.2 The Quadratic Formula","rendered":"6.2 The Quadratic Formula"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve quadratic equations using the quadratic formula for equations with integer, rational, irrational, or complex number solutions.<\/li>\r\n \t<li>Find [latex]x[\/latex]-intercepts (if they exist) of quadratic functions using the quadratic formula.<\/li>\r\n<\/ul>\r\n<\/div>\r\nYou can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for [latex]x[\/latex], you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.\r\n\r\nWe can derive the quadratic formula by completing the square. First,\u00a0assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:\r\n<ol>\r\n \t<li>First, move the constant term to the right side of the equal sign:\r\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\r\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div><\/li>\r\n \t<li>Now, use the square root property, which gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div><\/li>\r\n<\/ol>\r\nHere is a video explanation of the steps given above.\r\n\r\nhttps:\/\/youtu.be\/AD58TWGIcsQ?si=xhau-nfPlgxR57zG&amp;t=3\r\n\r\nThis formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].\r\n\r\nThe form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it is\u00a0<i>vital<\/i> that you be sure the equation is in this form. If you do not, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.\r\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\r\nThe Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.\r\n<ol>\r\n \t<li>Put the equation in standard form first.<\/li>\r\n \t<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be sure to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\r\n \t<li>Carefully substitute the values noted in step\u00a0[latex]2[\/latex] into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\r\n \t<li>Simplify as much as possible.<\/li>\r\n \t<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added and one in which it is subtracted<i>.<\/i><\/li>\r\n \t<li>Simplify both values to get the possible solutions.<\/li>\r\n<\/ol>\r\nThat is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will go back and solve it again using another factoring method.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].\r\n\r\n[reveal-answer q=\"296770\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296770\"]\r\n\r\nFirst write the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1, b=4, c=-5\\end{array}[\/latex]<\/p>\r\nNote that the subtraction sign means the constant <i>c<\/i> is negative.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\u00a0[latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify, being careful to get the signs correct.\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\r\nSimplify some more.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\r\nSimplify the radical: [latex] \\sqrt{36}=6[\/latex].\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\r\nSeparate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other,\u00a0[latex]6[\/latex] is subtracted.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nYou can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nYou get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You have solved the equation successfully using the Quadratic Formula!\r\n\r\nWatch this video to see an example of how to use the quadratic formula to solve a quadratic equation that has two real, rational solutions.\r\n\r\nhttps:\/\/youtu.be\/xtwO-n8lRPw\r\n\r\nSometimes, it may be easier to solve an equation using conventional factoring methods like finding number pairs that sum to one number (in this example, [latex]4[\/latex]) and that produce a specific product (in this example [latex]\u22125[\/latex]) when multiplied. The power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.\r\n\r\nIn the next video example, we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.\r\n\r\nhttps:\/\/youtu.be\/tF0muV86dr0\r\n\r\nMost of the quadratic equations you have looked at have two solutions, like the one above. The following example is a little different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].\r\n\r\n[reveal-answer q=\"998241\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"998241\"]\r\n\r\nSubtract [latex]6[\/latex]<i>x <\/i>from each side and add\u00a0[latex]16[\/latex] to both sides to put the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\r\nIdentify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\r\nSince the square root of\u00a0[latex]0[\/latex] is\u00a0[latex]0[\/latex], and both adding and subtracting\u00a0[latex]0[\/latex] give the same result, there is only one possible value.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\r\nThe answer is [latex]x=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAgain, check using the original equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nIn the next example, we will show that some quadratic equations do not have real solutions. \u00a0As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[\/latex]\r\n[reveal-answer q=\"268005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"268005\"]\r\n\r\nAdd [latex]x[\/latex] to both sides and add 3 to both sides to get the quadratic equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+x=-x-3\\\\x^{2}+2x+3=0\\end{array}[\/latex]<\/p>\r\nIdentify a, b, c.\r\n<p style=\"text-align: center;\">[latex]a=1, b=2, c=3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute values for a, b, c into the quadratic formula.<\/p>\r\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Simplify<\/p>\r\n<p style=\"text-align: left;\">[latex] x=\\frac{-2\\pm \\sqrt{-8}}{2}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe have seen two outcomes for solutions to\u00a0quadratic equations; either there was one or two real number solutions. We have also learned that it is possible to take the square root of a negative number by using imaginary numbers. Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:\r\n<p style=\"text-align: center;\">[latex]2x^2+3x+6=0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Using the Quadratic Formula to solve this equation, we first identify a, b, and c.<\/p>\r\n<p style=\"text-align: center;\">[latex]a = 2,b = 3,c = 6[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We can place a, b and c into the quadratic formula and simplify to get the following result:<\/p>\r\n<p style=\"text-align: center;\">[latex]x=-\\frac{3}{4}+\\frac{\\sqrt{-39}}{4}, x=-\\frac{3}{4}-\\frac{\\sqrt{-39}}{4}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Up to this point, we would have said\u00a0that [latex]\\sqrt{-39}[\/latex] is not defined for real numbers and determine that this equation has no solutions. \u00a0But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.<\/p>\r\n<p style=\"text-align: center;\">[latex]x=-\\frac{3}{4}+i\\frac{\\sqrt{39}}{4}, x=-\\frac{3}{4}-i\\frac{\\sqrt{39}}{4}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">In this section we will practice simplifying and writing solutions to quadratic equations that are complex. \u00a0We will then present a technique for classifying whether the solution(s) to a quadratic equation will be complex, and how many solutions there will be.<\/p>\r\n<p style=\"text-align: left;\">In our first example, we will work through the process of solving\u00a0a quadratic equation with\u00a0complex solutions. Take note that we will be simplifying complex numbers, so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.<\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}+2x=-5[\/latex].\r\n\r\n[reveal-answer q=\"654640\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"654640\"]\r\n\r\nFirst write the equation in standard form.\r\n\r\n[latex]\\begin{array}{l}x^{2}+2x=-5\\\\x^{2}+2x+5=0\\,\\,\\,\\,\\,\\\\\\\\a=1,b=2,c=5\\,\\,\\,\\end{array}[\/latex]\r\n\r\n[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,2x\\,\\,\\,+\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]\r\n\r\nSubstitute the values into the Quadratic Formula.\r\n\r\n[latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(5)}}{2(1)}[\/latex]\r\n\r\nSimplify, being careful to get the signs correct.\r\n\r\n[latex] x=\\frac{-2\\pm \\sqrt{4-20}}{2}[\/latex]\r\n\r\nSimplify some more.\r\n\r\n[latex] x=\\frac{-2\\pm \\sqrt{-16}}{2}[\/latex]\r\n\r\nSimplify the radical, but notice that the number under the radical symbol is negative! The square root of [latex]\u221216[\/latex] is imaginary. [latex] \\sqrt{-16}=4i[\/latex].\r\n\r\n[latex] x=\\frac{-2\\pm 4i}{2}[\/latex]\r\n\r\nSeparate and simplify to find the solutions to the quadratic equation.\r\n\r\n[latex]\\begin{array}{c}x=\\frac{-2+4i}{2}=\\frac{-2}{2}+\\frac{4i}{2}=-1+2i\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-2-4i}{2}=\\frac{-2}{2}-\\frac{2i}{4}\\cdot \\frac{2}{2}=-1-2i\\end{array}[\/latex]\r\n\r\nTherefore, [latex]x=-1+2i[\/latex] or [latex]-1-2i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can check these solutions in the original equation. Be careful when you expand the squares, and replace [latex]i^{2}[\/latex]\u00a0with [latex]-1[\/latex].\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{r}x=-1+2i\\\\x^{2}+2x=-5\\\\\\left(-1+2i\\right)^{2}+2\\left(-1+2i\\right)=-5\\\\1-4i+4i^{2}-2+4i=-5\\\\1-4i+4\\left(-1\\right)-2+4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{r}x=-1-2i\\\\x^{2}+2x=-5\\\\\\left(-1-2i\\right)^{2}+2\\left(-1-2i\\right)=-5\\\\1+4i+4i^{2}-2-4i=-5\\\\1+4i+4\\left(-1\\right)-2-4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165135500790\">Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\r\n[reveal-answer q=\"144216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"144216\"]\r\n\r\nFirst, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex]. Substitute these values into the quadratic formula. [latex]\\begin{array}{l}x\\hfill&amp;=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&amp;=\\frac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&amp;=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&amp;=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&amp;=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]\r\n\r\nNow we can separate the expression [latex]\\frac{-1\\pm i\\sqrt{7}}{2}[\/latex] into two solutions:\r\n\r\n[latex]-\\frac{1}{2}+\\frac{ i\\sqrt{7}}{2}[\/latex]\r\n\r\n[latex]-\\frac{1}{2}-\\frac{ i\\sqrt{7}}{2}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThe solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex]. It is important that you separate the real and imaginary part as this is proper complex number form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow that we have had a little practice solving quadratic equations whose solutions are complex, we can explore a related\u00a0feature of quadratic functions. Consider the following function: [latex]f(x)=x^2+2x+3[\/latex]. \u00a0Recall that the x-intercepts of a function are found by setting the function equal to zero:\r\n<p style=\"text-align: center;\">[latex]x^2+2x+3=0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The function now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the original function and see that it has no x-intercepts.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the x-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[\/latex]\r\n[reveal-answer q=\"698410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698410\"]\r\n\r\nThe x-intercepts of the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] are found by setting it equal to zero and solving for x since the y values of the x-intercepts are zero.\r\n\r\nFirst, identify a, b, c.\r\n\r\n[latex]\\begin{array}{l}x^2+2x+3=0\\\\a=1,b=2,c=3\\end{array}[\/latex]\r\n\r\nSubstitute these values into the quadratic formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}x &amp; =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\ &amp; =\\frac{-2\\pm \\sqrt{{2}^{2}-4(1)(3)}}{2(1)} \\\\ &amp; =\\frac{-2\\pm \\sqrt{4-12}}{2} \\\\ &amp; =\\frac{-2\\pm \\sqrt{-8}}{2} \\\\ &amp; =\\frac{-2\\pm 2i\\sqrt{2}}{2} \\\\ &amp; =-1\\pm i\\sqrt{2} \\\\ &amp;=-1+i\\sqrt{2},-1-i\\sqrt{2}\\end{array}[\/latex]<\/p>\r\nThe solutions to this equations are complex, therefore there are no x-intercepts for the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:\r\n\r\n[caption id=\"attachment_3475\" align=\"aligncenter\" width=\"241\"]<img class=\"wp-image-3475\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/04190045\/Screen-Shot-2016-08-04-at-11.34.19-AM.png\" alt=\"Graph of quadratic function with the following points (-1,2), (-2,3), (0,3), (1,6), (-3,6). \" width=\"241\" height=\"249\" \/> Graph of quadratic function with no x-intercepts in the real numbers.[\/caption]\r\n\r\nNote how the graph does not cross the x-axis; therefore, there are no real x-intercepts for this function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video gives another example of how to use the quadratic formula to find solutions to a quadratic equation that has complex solutions.\r\n\r\nhttps:\/\/youtu.be\/11EwTcRMPn8\r\n<h2>Summary<\/h2>\r\nQuadratic equations can have complex solutions. Quadratic functions whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]f(x)=0[\/latex].\r\n<h2>Applying the Quadratic Formula<\/h2>\r\nQuadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.\r\n\r\nA very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation is not completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA ball is thrown off a building from\u00a0[latex]200[\/latex] feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. The negative value means it is heading toward the ground.\r\n\r\nThe equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after <i>t<\/i> seconds. About how long does it take for the ball to hit the ground?\r\n\r\n[reveal-answer q=\"704677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704677\"]\r\n\r\nWhen the ball hits the ground, the height is\u00a0[latex]0[\/latex]. Substitute\u00a0[latex]0[\/latex] for <i>h<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\r\nThis equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is <i>t<\/i> rather than <i>x<\/i>. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].\r\n<p style=\"text-align: center;\">[latex] t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\r\nSimplify. Be very careful with the signs.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\r\nUse a calculator to find both roots.\r\n<p style=\"text-align: center;\"><i>t<\/i> is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\r\nConsider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.\r\n\r\nThe ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.\r\n\r\nhttps:\/\/youtu.be\/RcVeuJhcuL0\r\n\r\nThe area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nBob made a quilt that is\u00a0[latex]4[\/latex] ft [latex]\\times[\/latex]\u00a0[latex]5[\/latex] ft. He has\u00a0[latex]10[\/latex] sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)\r\n\r\n[reveal-answer q=\"932211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932211\"]\r\n\r\nSketch the problem. Since you do not know the width of the border, you will let the variable [latex]x[\/latex] represent the width.\r\n\r\nIn the diagram. The original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/>\r\n\r\nSince each side of the original\u00a0[latex]4[\/latex] by\u00a0[latex]5[\/latex] quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].\r\n\r\n(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/>\r\n\r\nYou are only interested in the area of the border strips. Write an expression for the area of the border.\r\n<p style=\"text-align: center;\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\r\n<p style=\"text-align: center;\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\r\nThere are\u00a0[latex]10[\/latex] sq ft of fabric for the border, so set the area of border to be\u00a0[latex]10[\/latex].\r\n<p style=\"text-align: center;\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\r\nMultiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]10=18x+4x^{2}[\/latex]<\/p>\r\nSubtract\u00a0[latex]10[\/latex] from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}+18x-10=0\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\r\nFactor out the greatest common factor,\u00a0[latex]2[\/latex], so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\r\nUse the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\r\nFind the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\r\nIgnore the solution [latex]x=\u22125[\/latex], since the width could not be negative.\r\n\r\nThe width of the border should be\u00a0[latex]0.5[\/latex] ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.\r\n\r\nhttps:\/\/youtu.be\/Zxe-SdwutxA\r\n<h2>Summary<\/h2>\r\nQuadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations or any other quadratic equation! The Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex] [\/latex].\u00a0When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.\r\n\r\n&nbsp;\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the quadratic formula to solve quadratic equations with complex solutions<\/li>\r\n \t<li>Connect complex solutions with the graph of a quadratic function that does not cross the x-axis<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have seen two outcomes for solutions to\u00a0quadratic equations; either there was one or two real number solutions. We have also learned that it is possible to take the square root of a negative number by using imaginary numbers. Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:\r\n\r\n[latex]2x^2+3x+6=0[\/latex]\r\n\r\nUsing the Quadratic Formula to solve this equation, we first identify a, b, and c.\r\n\r\n[latex]a = 2,b = 3,c = 6[\/latex]\r\n\r\nWe can place a, b and c into the quadratic formula and simplify to get the following result:\r\n\r\n[latex]x=-\\frac{3}{4}+\\frac{\\sqrt{-39}}{4}, x=-\\frac{3}{4}-\\frac{\\sqrt{-39}}{4}[\/latex]\r\n\r\nUp to this point, we would have said\u00a0that [latex]\\sqrt{-39}[\/latex] is not defined for real numbers and determine that this equation has no solutions. \u00a0But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.\r\n\r\n[latex]x=-\\frac{3}{4}+i\\frac{\\sqrt{39}}{4}, x=-\\frac{3}{4}-i\\frac{\\sqrt{39}}{4}[\/latex]\r\n\r\nIn this section we will practice simplifying and writing solutions to quadratic equations that are complex. \u00a0We will then present a technique for classifying whether the solution(s) to a quadratic equation will be complex, and how many solutions there will be.\r\n\r\nIn our first example, we will work through the process of solving\u00a0a quadratic equation with\u00a0complex solutions. Take note that we will be simplifying complex numbers, so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}+2x=-5[\/latex].\r\n\r\n[reveal-answer q=\"654640\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"654640\"]\r\n\r\nFirst write the equation in standard form.\r\n\r\n[latex]\\begin{array}{l}x^{2}+2x=-5\\\\x^{2}+2x+5=0\\,\\,\\,\\,\\,\\\\\\\\a=1,b=2,c=5\\,\\,\\,\\end{array}[\/latex]\r\n\r\n[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,2x\\,\\,\\,+\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]\r\n\r\nSubstitute the values into the Quadratic Formula.\r\n\r\n[latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(5)}}{2(1)}[\/latex]\r\n\r\nSimplify, being careful to get the signs correct.\r\n\r\n[latex] x=\\frac{-2\\pm \\sqrt{4-20}}{2}[\/latex]\r\n\r\nSimplify some more.\r\n\r\n[latex] x=\\frac{-2\\pm \\sqrt{-16}}{2}[\/latex]\r\n\r\nSimplify the radical, but notice that the number under the radical symbol is negative! The square root of [latex]\u221216[\/latex] is imaginary. [latex] \\sqrt{-16}=4i[\/latex].\r\n\r\n[latex] x=\\frac{-2\\pm 4i}{2}[\/latex]\r\n\r\nSeparate and simplify to find the solutions to the quadratic equation.\r\n\r\n[latex]\\begin{array}{c}x=\\frac{-2+4i}{2}=\\frac{-2}{2}+\\frac{4i}{2}=-1+2i\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-2-4i}{2}=\\frac{-2}{2}-\\frac{2i}{4}\\cdot \\frac{2}{2}=-1-2i\\end{array}[\/latex]\r\n\r\nTherefore, [latex]x=-1+2i[\/latex] or [latex]-1-2i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can check these solutions in the original equation. Be careful when you expand the squares, and replace [latex]i^{2}[\/latex]\u00a0with [latex]-1[\/latex].\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{r}x=-1+2i\\\\x^{2}+2x=-5\\\\\\left(-1+2i\\right)^{2}+2\\left(-1+2i\\right)=-5\\\\1-4i+4i^{2}-2+4i=-5\\\\1-4i+4\\left(-1\\right)-2+4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{r}x=-1-2i\\\\x^{2}+2x=-5\\\\\\left(-1-2i\\right)^{2}+2\\left(-1-2i\\right)=-5\\\\1+4i+4i^{2}-2-4i=-5\\\\1+4i+4\\left(-1\\right)-2-4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165135500790\">Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\r\n[reveal-answer q=\"144216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"144216\"]\r\n\r\nFirst, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex]. Substitute these values into the quadratic formula. [latex]\\begin{array}{l}x\\hfill&amp;=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&amp;=\\frac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&amp;=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&amp;=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&amp;=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]\r\n\r\nNow we can separate the expression [latex]\\frac{-1\\pm i\\sqrt{7}}{2}[\/latex] into two solutions:\r\n\r\n[latex]-\\frac{1}{2}+\\frac{ i\\sqrt{7}}{2}[\/latex]\r\n\r\n[latex]-\\frac{1}{2}-\\frac{ i\\sqrt{7}}{2}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThe solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex]. It is important that you separate the real and imaginary part as this is proper complex number form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow that we have had a little practice solving quadratic equations whose solutions are complex, we can explore a related\u00a0feature of quadratic functions. Consider the following function: [latex]f(x)=x^2+2x+3[\/latex]. \u00a0Recall that the x-intercepts of a function are found by setting the function equal to zero:\r\n\r\n[latex]x^2+2x+3=0[\/latex]\r\n\r\nThe function now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the original function and see that it has no x-intercepts.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the x-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[\/latex]\r\n[reveal-answer q=\"698410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698410\"]\r\n\r\nThe x-intercepts of the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] are found by setting it equal to zero and solving for x since the y values of the x-intercepts are zero.\r\n\r\nFirst, identify a, b, c.\r\n\r\n[latex]\\begin{array}{l}x^2+2x+3=0\\\\a=1,b=2,c=3\\end{array}[\/latex]\r\n\r\nSubstitute these values into the quadratic formula.\r\n\r\n[latex]\\begin{array}{ll}x &amp; =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\ &amp; =\\frac{-2\\pm \\sqrt{{2}^{2}-4(1)(3)}}{2(1)} \\\\ &amp; =\\frac{-2\\pm \\sqrt{4-12}}{2} \\\\ &amp; =\\frac{-2\\pm \\sqrt{-8}}{2} \\\\ &amp; =\\frac{-2\\pm 2i\\sqrt{2}}{2} \\\\ &amp; =-1\\pm i\\sqrt{2} \\\\ &amp;=-1+i\\sqrt{2},-1-i\\sqrt{2}\\end{array}[\/latex]\r\n\r\nThe solutions to this equations are complex, therefore there are no x-intercepts for the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:\r\n\r\n[caption id=\"attachment_3475\" align=\"aligncenter\" width=\"241\"]<img class=\"wp-image-3475\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/04190045\/Screen-Shot-2016-08-04-at-11.34.19-AM.png\" alt=\"Graph of quadratic function with the following points (-1,2), (-2,3), (0,3), (1,6), (-3,6). \" width=\"241\" height=\"249\" \/> Graph of quadratic function with no x-intercepts in the real numbers.[\/caption]\r\n\r\nNote how the graph does not cross the x-axis; therefore, there are no real x-intercepts for this function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video gives another example of how to use the quadratic formula to find solutions to a quadratic equation that has complex solutions.\r\n\r\nhttps:\/\/youtu.be\/11EwTcRMPn8\r\n<h2>Summary<\/h2>\r\nQuadratic equations can have complex solutions. Quadratic functions whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]f(x)=0[\/latex].","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve quadratic equations using the quadratic formula for equations with integer, rational, irrational, or complex number solutions.<\/li>\n<li>Find [latex]x[\/latex]-intercepts (if they exist) of quadratic functions using the quadratic formula.<\/li>\n<\/ul>\n<\/div>\n<p>You can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for [latex]x[\/latex], you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.<\/p>\n<p>We can derive the quadratic formula by completing the square. First,\u00a0assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:<\/p>\n<ol>\n<li>First, move the constant term to the right side of the equal sign:\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div>\n<\/li>\n<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div>\n<\/li>\n<li>Now, use the square root property, which gives\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>Here is a video explanation of the steps given above.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"You use the quadratic formula all the time, but where did it come from?\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/AD58TWGIcsQ?start=3&#38;feature=oembed\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].<\/p>\n<p>The form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it is\u00a0<i>vital<\/i> that you be sure the equation is in this form. If you do not, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.<\/p>\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\n<p>The Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.<\/p>\n<ol>\n<li>Put the equation in standard form first.<\/li>\n<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be sure to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\n<li>Carefully substitute the values noted in step\u00a0[latex]2[\/latex] into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\n<li>Simplify as much as possible.<\/li>\n<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added and one in which it is subtracted<i>.<\/i><\/li>\n<li>Simplify both values to get the possible solutions.<\/li>\n<\/ol>\n<p>That is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will go back and solve it again using another factoring method.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q296770\">Show Solution<\/span><\/p>\n<div id=\"q296770\" class=\"hidden-answer\" style=\"display: none\">\n<p>First write the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1, b=4, c=-5\\end{array}[\/latex]<\/p>\n<p>Note that the subtraction sign means the constant <i>c<\/i> is negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.\u00a0[latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify, being careful to get the signs correct.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\n<p>Simplify some more.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\n<p>Simplify the radical: [latex]\\sqrt{36}=6[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\n<p>Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other,\u00a0[latex]6[\/latex] is subtracted.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>You can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>You get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You have solved the equation successfully using the Quadratic Formula!<\/p>\n<p>Watch this video to see an example of how to use the quadratic formula to solve a quadratic equation that has two real, rational solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Quadratic Formula - Two Real Rational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xtwO-n8lRPw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes, it may be easier to solve an equation using conventional factoring methods like finding number pairs that sum to one number (in this example, [latex]4[\/latex]) and that produce a specific product (in this example [latex]\u22125[\/latex]) when multiplied. The power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.<\/p>\n<p>In the next video example, we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex2:  Quadratic Formula - Two Real Irrational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tF0muV86dr0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Most of the quadratic equations you have looked at have two solutions, like the one above. The following example is a little different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q998241\">Show Solution<\/span><\/p>\n<div id=\"q998241\" class=\"hidden-answer\" style=\"display: none\">\n<p>Subtract [latex]6[\/latex]<i>x <\/i>from each side and add\u00a0[latex]16[\/latex] to both sides to put the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\n<p>Identify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\n<p>Since the square root of\u00a0[latex]0[\/latex] is\u00a0[latex]0[\/latex], and both adding and subtracting\u00a0[latex]0[\/latex] give the same result, there is only one possible value.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\n<p>The answer is [latex]x=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Again, check using the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>In the next example, we will show that some quadratic equations do not have real solutions. \u00a0As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q268005\">Show Solution<\/span><\/p>\n<div id=\"q268005\" class=\"hidden-answer\" style=\"display: none\">\n<p>Add [latex]x[\/latex] to both sides and add 3 to both sides to get the quadratic equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+x=-x-3\\\\x^{2}+2x+3=0\\end{array}[\/latex]<\/p>\n<p>Identify a, b, c.<\/p>\n<p style=\"text-align: center;\">[latex]a=1, b=2, c=3[\/latex]<\/p>\n<p style=\"text-align: left;\">Substitute values for a, b, c into the quadratic formula.<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify<\/p>\n<p style=\"text-align: left;\">[latex]x=\\frac{-2\\pm \\sqrt{-8}}{2}[\/latex]<\/p>\n<p style=\"text-align: left;\">Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We have seen two outcomes for solutions to\u00a0quadratic equations; either there was one or two real number solutions. We have also learned that it is possible to take the square root of a negative number by using imaginary numbers. Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:<\/p>\n<p style=\"text-align: center;\">[latex]2x^2+3x+6=0[\/latex]<\/p>\n<p style=\"text-align: left;\">Using the Quadratic Formula to solve this equation, we first identify a, b, and c.<\/p>\n<p style=\"text-align: center;\">[latex]a = 2,b = 3,c = 6[\/latex]<\/p>\n<p style=\"text-align: left;\">We can place a, b and c into the quadratic formula and simplify to get the following result:<\/p>\n<p style=\"text-align: center;\">[latex]x=-\\frac{3}{4}+\\frac{\\sqrt{-39}}{4}, x=-\\frac{3}{4}-\\frac{\\sqrt{-39}}{4}[\/latex]<\/p>\n<p style=\"text-align: left;\">Up to this point, we would have said\u00a0that [latex]\\sqrt{-39}[\/latex] is not defined for real numbers and determine that this equation has no solutions. \u00a0But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.<\/p>\n<p style=\"text-align: center;\">[latex]x=-\\frac{3}{4}+i\\frac{\\sqrt{39}}{4}, x=-\\frac{3}{4}-i\\frac{\\sqrt{39}}{4}[\/latex]<\/p>\n<p style=\"text-align: left;\">In this section we will practice simplifying and writing solutions to quadratic equations that are complex. \u00a0We will then present a technique for classifying whether the solution(s) to a quadratic equation will be complex, and how many solutions there will be.<\/p>\n<p style=\"text-align: left;\">In our first example, we will work through the process of solving\u00a0a quadratic equation with\u00a0complex solutions. Take note that we will be simplifying complex numbers, so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}+2x=-5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q654640\">Show Solution<\/span><\/p>\n<div id=\"q654640\" class=\"hidden-answer\" style=\"display: none\">\n<p>First write the equation in standard form.<\/p>\n<p>[latex]\\begin{array}{l}x^{2}+2x=-5\\\\x^{2}+2x+5=0\\,\\,\\,\\,\\,\\\\\\\\a=1,b=2,c=5\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,2x\\,\\,\\,+\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.<\/p>\n<p>[latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(5)}}{2(1)}[\/latex]<\/p>\n<p>Simplify, being careful to get the signs correct.<\/p>\n<p>[latex]x=\\frac{-2\\pm \\sqrt{4-20}}{2}[\/latex]<\/p>\n<p>Simplify some more.<\/p>\n<p>[latex]x=\\frac{-2\\pm \\sqrt{-16}}{2}[\/latex]<\/p>\n<p>Simplify the radical, but notice that the number under the radical symbol is negative! The square root of [latex]\u221216[\/latex] is imaginary. [latex]\\sqrt{-16}=4i[\/latex].<\/p>\n<p>[latex]x=\\frac{-2\\pm 4i}{2}[\/latex]<\/p>\n<p>Separate and simplify to find the solutions to the quadratic equation.<\/p>\n<p>[latex]\\begin{array}{c}x=\\frac{-2+4i}{2}=\\frac{-2}{2}+\\frac{4i}{2}=-1+2i\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-2-4i}{2}=\\frac{-2}{2}-\\frac{2i}{4}\\cdot \\frac{2}{2}=-1-2i\\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]x=-1+2i[\/latex] or [latex]-1-2i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can check these solutions in the original equation. Be careful when you expand the squares, and replace [latex]i^{2}[\/latex]\u00a0with [latex]-1[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{r}x=-1+2i\\\\x^{2}+2x=-5\\\\\\left(-1+2i\\right)^{2}+2\\left(-1+2i\\right)=-5\\\\1-4i+4i^{2}-2+4i=-5\\\\1-4i+4\\left(-1\\right)-2+4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{r}x=-1-2i\\\\x^{2}+2x=-5\\\\\\left(-1-2i\\right)^{2}+2\\left(-1-2i\\right)=-5\\\\1+4i+4i^{2}-2-4i=-5\\\\1+4i+4\\left(-1\\right)-2-4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165135500790\">Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q144216\">Show Solution<\/span><\/p>\n<div id=\"q144216\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex]. Substitute these values into the quadratic formula. [latex]\\begin{array}{l}x\\hfill&=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&=\\frac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Now we can separate the expression [latex]\\frac{-1\\pm i\\sqrt{7}}{2}[\/latex] into two solutions:<\/p>\n<p>[latex]-\\frac{1}{2}+\\frac{ i\\sqrt{7}}{2}[\/latex]<\/p>\n<p>[latex]-\\frac{1}{2}-\\frac{ i\\sqrt{7}}{2}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex]. It is important that you separate the real and imaginary part as this is proper complex number form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now that we have had a little practice solving quadratic equations whose solutions are complex, we can explore a related\u00a0feature of quadratic functions. Consider the following function: [latex]f(x)=x^2+2x+3[\/latex]. \u00a0Recall that the x-intercepts of a function are found by setting the function equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]x^2+2x+3=0[\/latex]<\/p>\n<p style=\"text-align: left;\">The function now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the original function and see that it has no x-intercepts.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the x-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698410\">Show Solution<\/span><\/p>\n<div id=\"q698410\" class=\"hidden-answer\" style=\"display: none\">\n<p>The x-intercepts of the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] are found by setting it equal to zero and solving for x since the y values of the x-intercepts are zero.<\/p>\n<p>First, identify a, b, c.<\/p>\n<p>[latex]\\begin{array}{l}x^2+2x+3=0\\\\a=1,b=2,c=3\\end{array}[\/latex]<\/p>\n<p>Substitute these values into the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}x & =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\ & =\\frac{-2\\pm \\sqrt{{2}^{2}-4(1)(3)}}{2(1)} \\\\ & =\\frac{-2\\pm \\sqrt{4-12}}{2} \\\\ & =\\frac{-2\\pm \\sqrt{-8}}{2} \\\\ & =\\frac{-2\\pm 2i\\sqrt{2}}{2} \\\\ & =-1\\pm i\\sqrt{2} \\\\ &=-1+i\\sqrt{2},-1-i\\sqrt{2}\\end{array}[\/latex]<\/p>\n<p>The solutions to this equations are complex, therefore there are no x-intercepts for the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:<\/p>\n<div id=\"attachment_3475\" style=\"width: 251px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3475\" class=\"wp-image-3475\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/04190045\/Screen-Shot-2016-08-04-at-11.34.19-AM.png\" alt=\"Graph of quadratic function with the following points (-1,2), (-2,3), (0,3), (1,6), (-3,6).\" width=\"241\" height=\"249\" \/><\/p>\n<p id=\"caption-attachment-3475\" class=\"wp-caption-text\">Graph of quadratic function with no x-intercepts in the real numbers.<\/p>\n<\/div>\n<p>Note how the graph does not cross the x-axis; therefore, there are no real x-intercepts for this function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video gives another example of how to use the quadratic formula to find solutions to a quadratic equation that has complex solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Quadratic Formula - Complex Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/11EwTcRMPn8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Quadratic equations can have complex solutions. Quadratic functions whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]f(x)=0[\/latex].<\/p>\n<h2>Applying the Quadratic Formula<\/h2>\n<p>Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.<\/p>\n<p>A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation is not completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A ball is thrown off a building from\u00a0[latex]200[\/latex] feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. The negative value means it is heading toward the ground.<\/p>\n<p>The equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after <i>t<\/i> seconds. About how long does it take for the ball to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704677\">Show Solution<\/span><\/p>\n<div id=\"q704677\" class=\"hidden-answer\" style=\"display: none\">\n<p>When the ball hits the ground, the height is\u00a0[latex]0[\/latex]. Substitute\u00a0[latex]0[\/latex] for <i>h<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\n<p>This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is <i>t<\/i> rather than <i>x<\/i>. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\n<p>Simplify. Be very careful with the signs.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\n<p>Use a calculator to find both roots.<\/p>\n<p style=\"text-align: center;\"><i>t<\/i> is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\n<p>Consider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.<\/p>\n<p>The ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Quadratic Formula Application - Time for an Object to Hit the Ground\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RcVeuJhcuL0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Bob made a quilt that is\u00a0[latex]4[\/latex] ft [latex]\\times[\/latex]\u00a0[latex]5[\/latex] ft. He has\u00a0[latex]10[\/latex] sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932211\">Show Solution<\/span><\/p>\n<div id=\"q932211\" class=\"hidden-answer\" style=\"display: none\">\n<p>Sketch the problem. Since you do not know the width of the border, you will let the variable [latex]x[\/latex] represent the width.<\/p>\n<p>In the diagram. The original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/><\/p>\n<p>Since each side of the original\u00a0[latex]4[\/latex] by\u00a0[latex]5[\/latex] quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].<\/p>\n<p>(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/><\/p>\n<p>You are only interested in the area of the border strips. Write an expression for the area of the border.<\/p>\n<p style=\"text-align: center;\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\n<p style=\"text-align: center;\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\n<p>There are\u00a0[latex]10[\/latex] sq ft of fabric for the border, so set the area of border to be\u00a0[latex]10[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\n<p>Multiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]10=18x+4x^{2}[\/latex]<\/p>\n<p>Subtract\u00a0[latex]10[\/latex] from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}+18x-10=0\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\n<p>Factor out the greatest common factor,\u00a0[latex]2[\/latex], so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\n<p>Use the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\n<p>Find the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\n<p>Ignore the solution [latex]x=\u22125[\/latex], since the width could not be negative.<\/p>\n<p>The width of the border should be\u00a0[latex]0.5[\/latex] ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Our last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Quadratic Formula Application - Determine the Width of a Border\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Zxe-SdwutxA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations or any other quadratic equation! The Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex][\/latex].\u00a0When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the quadratic formula to solve quadratic equations with complex solutions<\/li>\n<li>Connect complex solutions with the graph of a quadratic function that does not cross the x-axis<\/li>\n<\/ul>\n<\/div>\n<p>We have seen two outcomes for solutions to\u00a0quadratic equations; either there was one or two real number solutions. We have also learned that it is possible to take the square root of a negative number by using imaginary numbers. Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:<\/p>\n<p>[latex]2x^2+3x+6=0[\/latex]<\/p>\n<p>Using the Quadratic Formula to solve this equation, we first identify a, b, and c.<\/p>\n<p>[latex]a = 2,b = 3,c = 6[\/latex]<\/p>\n<p>We can place a, b and c into the quadratic formula and simplify to get the following result:<\/p>\n<p>[latex]x=-\\frac{3}{4}+\\frac{\\sqrt{-39}}{4}, x=-\\frac{3}{4}-\\frac{\\sqrt{-39}}{4}[\/latex]<\/p>\n<p>Up to this point, we would have said\u00a0that [latex]\\sqrt{-39}[\/latex] is not defined for real numbers and determine that this equation has no solutions. \u00a0But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.<\/p>\n<p>[latex]x=-\\frac{3}{4}+i\\frac{\\sqrt{39}}{4}, x=-\\frac{3}{4}-i\\frac{\\sqrt{39}}{4}[\/latex]<\/p>\n<p>In this section we will practice simplifying and writing solutions to quadratic equations that are complex. \u00a0We will then present a technique for classifying whether the solution(s) to a quadratic equation will be complex, and how many solutions there will be.<\/p>\n<p>In our first example, we will work through the process of solving\u00a0a quadratic equation with\u00a0complex solutions. Take note that we will be simplifying complex numbers, so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}+2x=-5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q654640\">Show Solution<\/span><\/p>\n<div id=\"q654640\" class=\"hidden-answer\" style=\"display: none\">\n<p>First write the equation in standard form.<\/p>\n<p>[latex]\\begin{array}{l}x^{2}+2x=-5\\\\x^{2}+2x+5=0\\,\\,\\,\\,\\,\\\\\\\\a=1,b=2,c=5\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,2x\\,\\,\\,+\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.<\/p>\n<p>[latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(5)}}{2(1)}[\/latex]<\/p>\n<p>Simplify, being careful to get the signs correct.<\/p>\n<p>[latex]x=\\frac{-2\\pm \\sqrt{4-20}}{2}[\/latex]<\/p>\n<p>Simplify some more.<\/p>\n<p>[latex]x=\\frac{-2\\pm \\sqrt{-16}}{2}[\/latex]<\/p>\n<p>Simplify the radical, but notice that the number under the radical symbol is negative! The square root of [latex]\u221216[\/latex] is imaginary. [latex]\\sqrt{-16}=4i[\/latex].<\/p>\n<p>[latex]x=\\frac{-2\\pm 4i}{2}[\/latex]<\/p>\n<p>Separate and simplify to find the solutions to the quadratic equation.<\/p>\n<p>[latex]\\begin{array}{c}x=\\frac{-2+4i}{2}=\\frac{-2}{2}+\\frac{4i}{2}=-1+2i\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-2-4i}{2}=\\frac{-2}{2}-\\frac{2i}{4}\\cdot \\frac{2}{2}=-1-2i\\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]x=-1+2i[\/latex] or [latex]-1-2i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We can check these solutions in the original equation. Be careful when you expand the squares, and replace [latex]i^{2}[\/latex]\u00a0with [latex]-1[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{r}x=-1+2i\\\\x^{2}+2x=-5\\\\\\left(-1+2i\\right)^{2}+2\\left(-1+2i\\right)=-5\\\\1-4i+4i^{2}-2+4i=-5\\\\1-4i+4\\left(-1\\right)-2+4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{r}x=-1-2i\\\\x^{2}+2x=-5\\\\\\left(-1-2i\\right)^{2}+2\\left(-1-2i\\right)=-5\\\\1+4i+4i^{2}-2-4i=-5\\\\1+4i+4\\left(-1\\right)-2-4i=-5\\\\1-4-2=-5\\\\-5=-5\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165135500790\">Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q144216\">Show Solution<\/span><\/p>\n<div id=\"q144216\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex]. Substitute these values into the quadratic formula. [latex]\\begin{array}{l}x\\hfill&=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\\\\\hfill&=\\frac{-\\left(1\\right)\\pm \\sqrt{{\\left(1\\right)}^{2}-\\left(4\\right)\\cdot \\left(1\\right)\\cdot \\left(2\\right)}}{2\\cdot 1}\\hfill \\\\\\hfill&=\\frac{-1\\pm \\sqrt{1 - 8}}{2}\\hfill \\\\ \\hfill&=\\frac{-1\\pm \\sqrt{-7}}{2}\\hfill \\\\\\hfill&=\\frac{-1\\pm i\\sqrt{7}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Now we can separate the expression [latex]\\frac{-1\\pm i\\sqrt{7}}{2}[\/latex] into two solutions:<\/p>\n<p>[latex]-\\frac{1}{2}+\\frac{ i\\sqrt{7}}{2}[\/latex]<\/p>\n<p>[latex]-\\frac{1}{2}-\\frac{ i\\sqrt{7}}{2}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The solutions to the equation are [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex]. It is important that you separate the real and imaginary part as this is proper complex number form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now that we have had a little practice solving quadratic equations whose solutions are complex, we can explore a related\u00a0feature of quadratic functions. Consider the following function: [latex]f(x)=x^2+2x+3[\/latex]. \u00a0Recall that the x-intercepts of a function are found by setting the function equal to zero:<\/p>\n<p>[latex]x^2+2x+3=0[\/latex]<\/p>\n<p>The function now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the original function and see that it has no x-intercepts.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the x-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698410\">Show Solution<\/span><\/p>\n<div id=\"q698410\" class=\"hidden-answer\" style=\"display: none\">\n<p>The x-intercepts of the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] are found by setting it equal to zero and solving for x since the y values of the x-intercepts are zero.<\/p>\n<p>First, identify a, b, c.<\/p>\n<p>[latex]\\begin{array}{l}x^2+2x+3=0\\\\a=1,b=2,c=3\\end{array}[\/latex]<\/p>\n<p>Substitute these values into the quadratic formula.<\/p>\n<p>[latex]\\begin{array}{ll}x & =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a} \\\\ & =\\frac{-2\\pm \\sqrt{{2}^{2}-4(1)(3)}}{2(1)} \\\\ & =\\frac{-2\\pm \\sqrt{4-12}}{2} \\\\ & =\\frac{-2\\pm \\sqrt{-8}}{2} \\\\ & =\\frac{-2\\pm 2i\\sqrt{2}}{2} \\\\ & =-1\\pm i\\sqrt{2} \\\\ &=-1+i\\sqrt{2},-1-i\\sqrt{2}\\end{array}[\/latex]<\/p>\n<p>The solutions to this equations are complex, therefore there are no x-intercepts for the function\u00a0[latex]f(x)=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:<\/p>\n<div id=\"attachment_3475\" style=\"width: 251px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3475\" class=\"wp-image-3475\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/04190045\/Screen-Shot-2016-08-04-at-11.34.19-AM.png\" alt=\"Graph of quadratic function with the following points (-1,2), (-2,3), (0,3), (1,6), (-3,6).\" width=\"241\" height=\"249\" \/><\/p>\n<p id=\"caption-attachment-3475\" class=\"wp-caption-text\">Graph of quadratic function with no x-intercepts in the real numbers.<\/p>\n<\/div>\n<p>Note how the graph does not cross the x-axis; therefore, there are no real x-intercepts for this function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video gives another example of how to use the quadratic formula to find solutions to a quadratic equation that has complex solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex:  Quadratic Formula - Complex Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/11EwTcRMPn8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Quadratic equations can have complex solutions. Quadratic functions whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]f(x)=0[\/latex].<\/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-204\">\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>Quadratic Formula Application - Time for an Object to Hit the Ground. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RcVeuJhcuL0\">https:\/\/youtu.be\/RcVeuJhcuL0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Quadratic Formula Application - Determine the Width of a Border. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Zxe-SdwutxA\">https:\/\/youtu.be\/Zxe-SdwutxA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex2: Quadratic Formula - Two Real Irrational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tF0muV86dr0\">https:\/\/youtu.be\/tF0muV86dr0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Quadratic Formula - Two Real Rational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xtwO-n8lRPw\">https:\/\/youtu.be\/xtwO-n8lRPw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex2: Quadratic Formula - 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