{"id":2733,"date":"2017-03-25T17:19:57","date_gmt":"2017-03-25T17:19:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/?post_type=chapter&p=2733"},"modified":"2017-03-28T05:02:53","modified_gmt":"2017-03-28T05:02:53","slug":"nah-nah-3","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/nah-nah-3\/","title":{"raw":"Using the Quadratic Formula","rendered":"Using the Quadratic Formula"},"content":{"raw":"The fourth method of solving a quadratic equation<\/strong> is by using the quadratic formula<\/strong>, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.\r\n\r\nWe can derive the quadratic formula by completing the square<\/strong>. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive 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
    \r\n \t
  1. First, move the constant term to the right side of the equal sign:\r\n
    [latex]a{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t
  2. As we want the leading coefficient to equal 1, divide through by a<\/em>:\r\n
    [latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t
  3. 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
    [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
  4. 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
    [latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div><\/li>\r\n \t
  5. Now, use the square root property, which gives\r\n
    [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
  6. Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\r\n
    [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n
    \r\n

    A General Note: The Quadratic Formula<\/h3>\r\nWritten in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], any quadratic equation can be solved using the quadratic formula<\/strong>:\r\n
    [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\r\nwhere a<\/em>, b<\/em>, and c<\/em> are real numbers and [latex]a\\ne 0[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    How To: Given a quadratic equation, solve it using the quadratic formula<\/h3>\r\n
      \r\n \t
    1. Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[\/latex].<\/li>\r\n \t
    2. Make note of the values of the coefficients and constant term, [latex]a,b[\/latex], and [latex]c[\/latex].<\/li>\r\n \t
    3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\r\n \t
    4. Calculate and solve.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      Example 9: Solve the Quadratic Equation Using the Quadratic Formula<\/h3>\r\nSolve the quadratic equation: [latex]{x}^{2}+5x+1=0[\/latex].\r\n\r\n<\/div>\r\n
      \r\n

      Solution<\/h3>\r\nIdentify the coefficients: [latex]a=1,b=5,c=1[\/latex]. Then use the quadratic formula.\r\n
      [latex]\\begin{array}{l}x\\hfill&=\\frac{-\\left(5\\right)\\pm \\sqrt{{\\left(5\\right)}^{2}-4\\left(1\\right)\\left(1\\right)}}{2\\left(1\\right)}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{25 - 4}}{2}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{21}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\nThe solution of the quadratic equation in the previous example\u00a0is a real number that contains the square root of 21. What if we solve an equation whose solution\u00a0contains the square root of a negative number? In that case, the solution would not be a real number, since\u00a0the square of\u00a0every real number is a positive number. In fact, the solution would be a\u00a0complex number with an imaginary part.\r\n\r\nThe following example and Try It exercise assume you are familiar with complex numbers. If you need to review complex numbers beforehand, study the Complex Numbers section included below.\r\n
      \r\n

      Example 10: Solving a Quadratic Equation with the Quadratic Formula<\/h3>\r\nUse the quadratic formula to solve [latex]{x}^{2}+x+2=0[\/latex].\r\n\r\n<\/div>\r\n
      \r\n

      Solution<\/h3>\r\nFirst, we identify the coefficients: [latex]a=1,b=1[\/latex], and [latex]c=2[\/latex].\r\n\r\nSubstitute these values into the quadratic formula.\r\n
      [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]<\/div>\r\nThe solutions to the equation are [latex]x=\\frac{-1+i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1-i\\sqrt{7}}{2}[\/latex] or [latex]x=\\frac{-1}{2}+\\frac{i\\sqrt{7}}{2}[\/latex] and [latex]x=\\frac{-1}{2}-\\frac{i\\sqrt{7}}{2}[\/latex].\r\n\r\n<\/div>\r\n
      \r\n

      Try It 8<\/h3>\r\nSolve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      <\/h2>\r\n

      Complex Numbers<\/h2>\r\n

      We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number<\/strong>. The imaginary number [latex]i[\/latex] is defined as the square root of negative 1.<\/p>\r\n\r\n

      [latex]\\sqrt{-1}=i\\\\[\/latex]<\/div>\r\n

      So, using properties of radicals,<\/p>\r\n\r\n

      [latex]{i}^{2}={\\left(\\sqrt{-1}\\right)}^{2}=-1\\\\[\/latex]<\/div>\r\n

      We can write the square root of any negative number as a multiple of i<\/em>. Consider the square root of \u201325.<\/p>\r\n\r\n

      [latex]\\begin{cases} \\sqrt{-25}=\\sqrt{25\\cdot \\left(-1\\right)}\\hfill \\\\ \\text{ }=\\sqrt{25}\\sqrt{-1}\\hfill \\\\ \\text{ }=5i\\hfill \\end{cases}\\\\[\/latex]<\/div>\r\n

      We use 5i\u00a0<\/em>and not [latex]-\\text{5}i[\/latex]\u00a0because the principal root of 25 is the positive root.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Showing Figure 1<\/b>[\/caption]\r\n

      A complex number<\/strong> is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a\u00a0<\/em>+ bi<\/em>\u00a0where a<\/em>\u00a0is the real part and bi<\/em>\u00a0is the imaginary part. For example, [latex]5+2i[\/latex] is a complex number. So, too, is [latex]3+4\\sqrt{3}i[\/latex].<\/p>\r\n

      Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative\u00a0real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.<\/p>\r\n\r\n

      \r\n

      A General Note: Imaginary and Complex Numbers<\/h3>\r\n

      A complex number<\/strong> is a number of the form [latex]a+bi\\\\[\/latex] where<\/p>\r\n\r\n