Practice 2 Solutions

Determine the value that will make the expression a perfect square trinomial. Show your work.

$x^{2} + b x + 49$

$a = 1, c = 49 b = 2 \sqrt{a c} b = 2 \sqrt{(1) (49)} = 2 \sqrt{49} = 2 (7)$

$b = 14$

$x^{2} + b x + \frac{4}{9}$

$a = 1, c = \frac{4}{9} b = 2 \sqrt{a c} b = 2 \sqrt{(1) (\frac{4}{9})} = 2 \sqrt{\frac{4}{9}} = 2 (\frac{2}{3})$

$b = \frac{4}{3}$

$x^{2} - 24 x + c$

$a = 1, b = - 24 c = {(\frac{b}{2})}^{2} = {(- \frac{24}{2})}^{2} = {(- 12)}^{2}$

$c = 144$

$x^{2} + 6 x + c$

$a = 1, b = 6 c = {(\frac{b}{2})}^{2} = {(\frac{6}{2})}^{2} = {(3)}^{2}$

$c = 9$

$x^{2} + b x + 225$

$a = 1, c = 225 b = 2 \sqrt{a c} = 2 \sqrt{(1) (225)} = 2 \sqrt{225} = 2 (15)$

$b = 30$

Solve the quadratic equations by completing the square.

Note

Remember to set up the equation as $a x^{2} + b x = c$ in the first step.

$2 x^{2} - 3 x - 2 = 0$

$\begin{array}{rcl} \frac{2}{2} x^{2} - \frac{3}{2} x - \frac{2}{2} & = & 0 \\ x^{2} - \frac{3}{2} x - 1 & = & 0 \\ x^{2} - \frac{3}{2} x & = & 1 \\ x^{2} - \frac{3}{2} x + {(- \frac{3}{2 \cdot 2})}^{2} & = & 1 + {(- \frac{3}{2 \cdot 2})}^{2} \\ x^{2} - \frac{3}{2} x + {(- \frac{3}{4})}^{2} & = & 1 + {(- \frac{3}{4})}^{2} \\ {(x - \frac{3}{4})}^{2} & = & 1 + \frac{9}{16} \\ {(x - \frac{3}{4})}^{2} & = & \frac{16}{16} + \frac{9}{16} \\ {(x - \frac{3}{4})}^{2} & = & \frac{25}{16} \\ \sqrt{{(x - \frac{3}{4})}^{2}} & = & \pm \sqrt{\frac{25}{16}} \\ x - \frac{3}{4} & = & \pm \frac{5}{4} \\ x & = & \frac{3}{4} \pm \frac{5}{4} \end{array}$

$x = - \frac{1}{2}, 2$

$x^{2} = 10 x - 14$

$\begin{array}{rcl} x^{2} - 10 x & = & - 14 \\ x^{2} - 10 x + {(- 5)}^{2} & = & - 14 + {(- 5)}^{2} \\ {(x - 5)}^{2} & = & - 14 + 25 \\ {(x - 5)}^{2} & = & 11 \\ \sqrt{{(x - 5)}^{2}} & = & \pm \sqrt{11} \\ x - 5 & = & \pm \sqrt{11} \end{array}$

$x = 5 \pm \sqrt{11}$

$3 x^{2} + 9 x + 5 = 0$

$\begin{array}{rcl} \frac{3}{3} x^{2} + \frac{9}{3} x + \frac{5}{3} & = & 0 \\ x^{2} + 3 x + \frac{5}{3} & = & 0 \\ x^{2} + 3 x & = & - \frac{5}{3} \\ x^{2} + 3 x + {(\frac{3}{2})}^{2} & = & - \frac{5}{3} + {(\frac{3}{2})}^{2} \\ {(x + \frac{3}{2})}^{2} & = & - \frac{5}{3} + \frac{9}{4} \\ {(x + \frac{3}{2})}^{2} & = & - \frac{20}{12} + \frac{27}{12} \\ {(x + \frac{3}{2})}^{2} & = & \frac{7}{12} \\ \sqrt{{(x + \frac{3}{2})}^{2}} & = & \pm \sqrt{\frac{7}{12}} \\ x + \frac{3}{2} & = & \pm \frac{\sqrt{7}}{2 \sqrt{3}} \\ x + \frac{3}{2} & = & \pm \frac{\sqrt{7}}{2 \sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3}}) \\ x + \frac{3}{2} & = & \pm \frac{\sqrt{21}}{6} \end{array}$

$x = - \frac{3}{2} \pm \frac{\sqrt{21}}{6} or x = \frac{- 9 \pm \sqrt{21}}{6}$

$x^{2} + 20 x + 94 = 0$

$\begin{array}{rcl} x^{2} + 20 x & = & - 94 \\ x^{2} + 20 x + {(10)}^{2} & = & - 94 + {(10)}^{2} \\ {(x + 10)}^{2} & = & - 94 + 100 \\ {(x + 10)}^{2} & = & 6 \\ \sqrt{{(x + 10)}^{2}} & = & \pm \sqrt{6} \\ x + 10 & = & \pm \sqrt{6} \end{array}$

$\begin{array}{rcl} x & = & - 10 \pm \sqrt{6} \end{array}$

Note

Remember to rationalize the denominator.

It is not necessary to write the solution as one fraction. However, in the next lesson, you will use a formula to write the solution as one fraction when the solutions are imaginary or irrational.

$x^{2} - 2 x = - 38$

$\begin{array}{rcl} x^{2} - 2 x + {(- 1)}^{2} & = & - 38 + {(- 1)}^{2} \\ {(x - 1)}^{2} & = & - 38 + 1 \\ {(x - 1)}^{2} & = & - 37 \\ \sqrt{{(x - 1)}^{2}} & = & \pm \sqrt{- 37} \\ x - 1 & = & \pm i \sqrt{37} \end{array}$

$\begin{array}{rcl} x & = & 1 \pm i \sqrt{37} \end{array}$