{"id":16449,"date":"2019-10-03T15:48:57","date_gmt":"2019-10-03T15:48:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-quadratic-equations-with-complex-solutions\/"},"modified":"2024-05-02T15:48:30","modified_gmt":"2024-05-02T15:48:30","slug":"read-quadratic-equations-with-complex-solutions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-quadratic-equations-with-complex-solutions\/","title":{"raw":"Quadratic Equations With Complex Solutions","rendered":"Quadratic Equations With Complex Solutions"},"content":{"raw":"<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 equation that does not cross the [latex]x[\/latex]-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<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 <a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/16-4-1-complex-numbers\/\">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<\/a>.<\/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\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]25644[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Connect complex solutions with the graph of a quadratic equation<\/h2>\r\nNow that we have had a little practice solving quadratic equations whose solutions are complex, we can explore a related\u00a0feature of quadratic equations in two variables. Consider the following equation in two variables: [latex]y=x^2+2x+3[\/latex]. \u00a0<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/5-3-1-solving-quadratic-equations-by-factoring\/?preview_id=16465&amp;preview_nonce=d137c610b3&amp;preview=true\">Recall that we showed earlier how the x-intercepts of an equation's graph are found by setting the y-value equal to zero<\/a>:\r\n<p style=\"text-align: center;\">[latex]0=x^2+2x+3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">This now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the equation 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 equation [latex]y=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 equation [latex]y=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+\\sqrt{2},-1-\\sqrt{2}\\end{array}[\/latex]<\/p>\r\nThe solutions to this equations are complex, therefore there are no x-intercepts for the equation [latex]y=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the equation 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 equation 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 equation.\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 equations of the form [latex]y=ax^2+bx+c[\/latex] whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]y=0[\/latex].","rendered":"<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 equation that does not cross the [latex]x[\/latex]-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 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 <a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/16-4-1-complex-numbers\/\">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<\/a>.<\/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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm25644\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25644&theme=oea&iframe_resize_id=ohm25644&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Connect complex solutions with the graph of a quadratic equation<\/h2>\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 equations in two variables. Consider the following equation in two variables: [latex]y=x^2+2x+3[\/latex]. \u00a0<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/5-3-1-solving-quadratic-equations-by-factoring\/?preview_id=16465&amp;preview_nonce=d137c610b3&amp;preview=true\">Recall that we showed earlier how the x-intercepts of an equation&#8217;s graph are found by setting the y-value equal to zero<\/a>:<\/p>\n<p style=\"text-align: center;\">[latex]0=x^2+2x+3[\/latex]<\/p>\n<p style=\"text-align: left;\">This now looks like the type of quadratic equations we have been solving. In the next example, we will solve this equation, then graph the equation 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 equation [latex]y=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 equation [latex]y=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+\\sqrt{2},-1-\\sqrt{2}\\end{array}[\/latex]<\/p>\n<p>The solutions to this equations are complex, therefore there are no x-intercepts for the equation [latex]y=x^2+2x+3[\/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the equation 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 equation 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 equation.<\/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-1\" 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 equations of the form [latex]y=ax^2+bx+c[\/latex] whose graphs\u00a0do not cross the x-axis will have complex solutions for [latex]y=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-16449\">\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>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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>Ex: Quadratic Formula - Complex Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/11EwTcRMPn8\">https:\/\/youtu.be\/11EwTcRMPn8<\/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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":169554,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Quadratic Formula - Complex Solutions\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/11EwTcRMPn8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\" http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"482a5524ebbe4ca596b63be2557597bf, 74424c30f2994d5881b6b2351807af13, ebed27ec79b4433683a3da150ab816fe","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16449","chapter","type-chapter","status-publish","hentry"],"part":16201,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16449","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16449\/revisions"}],"predecessor-version":[{"id":19984,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16449\/revisions\/19984"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/16201"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16449\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16449"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16449"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16449"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16449"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}