Compare Your Answers
  1. A quadratic equation is given as \(36a^2 - 22 = 3\).

    1. Use the quadratic formula to solve the equation.

      \(\begin{array}{l}
       36a^2 - 22 = 3 \\
       36a^2 - 25 = 0 \\
       \end{array}\)

      \(a = 36, b = 0, c = -25\)

      \[\begin{array}
       x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
       x = \frac{{ - 0 \pm \sqrt {0^2 - 4\left( {36} \right)\left( { - 25} \right)} }}{{2\left( {36} \right)}} \\
       x = \frac{{ \pm \sqrt {3600} }}{{72}} \\
       x = \frac{{ \pm 60}}{{72}} \\
       x = \pm \frac{5}{6} \\
       x = \frac{5}{6} \thinspace {\rm{and} } \thinspace x = - \frac{5}{6}
       \end{array}\]

      Left Side
      		Right Side
      \[\begin{array}{r}
       36a^2 - 22 \\
       36\left( {\frac{5}{6}} \right)^2 - 22 \\
       36\left( {\frac{{25}}{{36}}} \right) - 22 \\
       25 - 22 \\
       3  \end{array}\]             		\(3\)
                  LS = RS

      Left Side
      		Right Side
      \[\begin{array}{r}
       36a^2 - 22 \\
       36\left( { - \frac{5}{6}} \right)^2 - 22 \\
       36\left( {\frac{{25}}{{36}}} \right) - 22 \\
       25 - 22 \\
       3  \end{array}\]             		\(3\)
                     LS = RS


    2. Explain how the equation could be solved by factoring.

      Rearrange the equation so one side is equal to zero. Factor by recognizing a difference of squares. Equate each factor to zero and solve for \(x\).
    3. Solve by factoring.

      \(\begin{align}
       36a^2 - 22 &= 3 \\
       36a^2 - 25 &= 0 \\
       \left( {6a - 5} \right)\left( {6a + 5} \right) &= 0 \\
       \end{align}\)

      \(\begin{align}
       6a - 5 &= 0 \\
       a &= \frac{5}{6}
       \end{align}\)
      \(\begin{align}
       6a + 5&= 0 \\
       a &= -\frac{5}{6}
       \end{align}\)


  2. Solve \(6x^2 + 5x + 12 = 4x^2 + 15\) using the quadratic formula.


    \(\begin{array}{l}
     6x^2 + 5x + 12 = 4x^2 + 15 \\
     2x^2 + 5x - 3 = 0 \\
     \end{array}\)

    \(a = 2, b = 5, c = -3\)

    \[\begin{align}
     x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
     x &= \frac{{ - 5 \pm \sqrt {5^2 - 4\left( 2 \right)\left( { - 3} \right)} }}{{2\left( 2 \right)}} \\
     x &= \frac{{ - 5 \pm \sqrt {25 + 24} }}{4} \\
     x &= \frac{{ - 5 \pm \sqrt {49} }}{4} \\
     x &= \frac{{ - 5 \pm 7}}{4} \end{align}\]

    \(x = \frac{{ - 5 + 7}}{4} = \frac{1}{2}\) and \(x = \frac{{ - 5 - 7}}{4} = -3 \)











    Notice that the original equation is used to verify. This helps to ensure no errors were made in the initial steps of the solution.

    Left Side
    		Right Side
    \[\begin{array}{r}
     6x^2 + 5x + 12 \\
     6\left( {\frac{1}{2}} \right)^2 + 5\left( {\frac{1}{2}} \right) + 12 \\
     6\left( {\frac{1}{4}} \right) + \frac{5}{2} + 12 \\
     \frac{6}{4} + \frac{{10}}{4} + \frac{{48}}{4} \\
     \frac{{64}}{4} \\
     16  \end{array}\]
    		\[\begin{array}{l}
     4x^2 + 15 \\
     4\left( {\frac{1}{2}} \right)^2 + 15 \\
     4\left( {\frac{1}{4}} \right) + 15 \\
     1 + 15 \\
     16  \end{array}\]
                      LS = RS

    Left Side
    		Right Side
    \[\begin{array}{r}
     6x^2 + 5x + 12 \\
     6\left( { - 3} \right)^2 + 5\left( { - 3} \right) + 12 \\
     6\left( 9 \right) - 15 + 12 \\
     54 - 15 + 12 \\
     51 \end{array}\]
    		 \[\begin{array}{l}
     4x^2 + 15 \\
     4\left( { - 3} \right)^2 + 15 \\
     4\left( 9 \right) + 15 \\
     36 + 15 \\
     51 \end{array}\]
                   LS = RS