Practice 1 Solutions - Demme Learning

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

$x^{2} + 16 x + c$

$a = 1, b = 16 c = {(\frac{b}{2})}^{2} = {(\frac{16}{2})}^{2} = 8^{2}$

$c = 64$

$x^{2} + 7 x + c$

$a = 1, b = 7 c = {(\frac{b}{2})}^{2} = {(\frac{7}{2})}^{2}$

$c = \frac{49}{4}$

$x^{2} + b x + 81$

$a = 1, c = 81 b = 2 \sqrt{a c} b = 2 \sqrt{(1) (81)} = 2 \sqrt{81} = 2 (9)$

$b = 18$

$x^{2} + b x + \frac{100}{81}$

$a = 1, c = \frac{100}{81} b = 2 \sqrt{a c} b = 2 \sqrt{(1) (\frac{100}{81})} = 2 \sqrt{\frac{100}{81}} = 2 (\frac{10}{9})$

$b = \frac{20}{9}$

$x^{2} + \frac{5}{2} x + c$

$a = 1, b = \frac{5}{2} c = {(\frac{b}{2})}^{2} = {(\frac{5}{2 \cdot 2})}^{2} = {(\frac{5}{4})}^{2}$

$c = \frac{25}{16}$

Solve the quadratic equations by completing the square.

$x^{2} - 6 x + 8 = 0$

$\begin{array}{rcl} x^{2} - 6 x & = & - 8 \\ x^{2} - 6 x + {(- 3)}^{2} & = & - 8 + {(- 3)}^{2} \\ {(x - 3)}^{2} & = & - 8 + 9 \\ {(x - 3)}^{2} & = & 1 \\ \sqrt{{(x - 3)}^{2}} & = & \pm \sqrt{1} \\ x - 3 & = & \pm 1 \\ x & = & 3 \pm 1 \\ x & = & 3 + 1 \\ x & = & 3 - 1 \end{array}$

$x = 2, 4$

$x^{2} - 2 x + 17 = 0$

$\begin{array}{rcl} x^{2} - 2 x & = & - 17 \\ x^{2} - 2 x + {(- 1)}^{2} & = & - 17 + {(- 1)}^{2} \\ {(x - 1)}^{2} & = & - 17 + 1 \\ {(x - 1)}^{2} & = & - 16 \\ \sqrt{{(x - 1)}^{2}} & = & \pm \sqrt{- 16} \\ x - 1 & = & \pm 4 i \end{array}$

$x = 1 \pm 4 i$

$x^{2} = 7 x + 4$

$\begin{array}{rcl} x^{2} - 7 x & = & 4 \\ x^{2} - 7 x + {(- \frac{7}{2})}^{2} & = & 4 + {(- \frac{7}{2})}^{2} \\ {(x - \frac{7}{2})}^{2} & = & 4 + \frac{49}{4} \\ {(x - \frac{7}{2})}^{2} & = & \frac{16}{4} + \frac{49}{4} \\ {(x - \frac{7}{2})}^{2} & = & \frac{65}{4} \\ \sqrt{{(x - \frac{7}{2})}^{2}} & = & \pm \sqrt{\frac{65}{4}} \\ x - \frac{7}{2} & = & \pm \frac{\sqrt{65}}{2} \end{array}$

$x = \frac{7 \pm \sqrt{65}}{2}$

Note

Remember to start with the equation in $a x^{2} + b x = c .$ This form allows you to write the left side of the equation as a perfect square trinomial.

$2 x^{2} + 8 x + 3 = 0$

$\begin{array}{rcl} \frac{2}{2} x^{2} + \frac{8}{2} x + \frac{3}{2} & = & 0 \\ x^{2} + 4 x + \frac{3}{2} & = & 0 \\ x^{2} + 4 x & = & - \frac{3}{2} \\ x^{2} + 4 x + {(2)}^{2} & = & - \frac{3}{2} + {(2)}^{2} \\ {(x + 2)}^{2} & = & - \frac{3}{2} + 4 \\ {(x + 2)}^{2} & = & - \frac{3}{2} + \frac{8}{2} \\ {(x + 2)}^{2} & = & \frac{5}{2} \\ \sqrt{{(x + 2)}^{2}} & = & \pm \sqrt{\frac{5}{2}} \\ x + 2 & = & \pm \frac{\sqrt{5}}{\sqrt{2}} (\frac{\sqrt{2}}{\sqrt{2}}) \\ x + 2 & = & \pm \frac{\sqrt{10}}{2} \end{array}$

$x = - 2 \pm \frac{\sqrt{10}}{2} or x = \frac{- 4 \pm \sqrt{10}}{2}$

Note

Q: What do you need to do when there is a square root in the denominator?

A: Rationalize the denominator.

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.

$4 x^{2} + 5 x + 4 = 0$

$\begin{array}{rcl} \frac{4}{4} x^{2} + \frac{5}{4} x + \frac{4}{4} & = & 0 \\ x^{2} + \frac{5}{4} x + 1 & = & 0 \\ x^{2} + \frac{5}{4} x & = & - 1 \\ x^{2} + \frac{5}{4} x + {(\frac{5}{2 \cdot 4})}^{2} & = & - 1 + {(\frac{5}{2 \cdot 4})}^{2} \\ x^{2} + \frac{5}{4} x + {(\frac{5}{8})}^{2} & = & - 1 + {(\frac{5}{8})}^{2} \\ {(x + \frac{5}{8})}^{2} & = & - 1 + \frac{25}{64} \\ {(x + \frac{5}{8})}^{2} & = & - \frac{64}{64} + \frac{25}{64} \\ {(x + \frac{5}{8})}^{2} & = & - \frac{39}{64} \\ \sqrt{{(x + \frac{5}{8})}^{2}} & = & \pm \sqrt{- \frac{39}{64}} \\ x + \frac{5}{8} & = & \pm \frac{i \sqrt{39}}{8} \end{array}$

$x = \frac{- 5 \pm i \sqrt{39}}{8}$