Chapter 9.4, Problem 19PPS

To determine

Solve the equation by completing the square method.

Answer to Problem 19PPS

The solutions are: − 8 and 2

Explanation of Solution

Given:

The equation: $x^2+6x-16=0$

Concept Used:

The perfect square term: ${(x\pm \frac{b}{2})}^2=x^2\pm 2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

Calculation:

The perfect square term: ${(x\pm \frac{b}{2})}^2=x^2\pm 2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

The equation: $x^2+6x-16=0$

  $\begin{array}{l}{x}^2+6x-16=0[\text{original}\text{equation}]\\ $ x^2+6x+9-16=0+9[\text{Since}{(\frac{6}{2})}^2=9\text{add}\text{to}\text{each}\text{side}]$x^2+6x+9-16+16=9+16[\text{add}\text{16}\text{to}\text{each}\text{side}]\\ $ {(x+3)}^2=25$\sqrt{{(x+3)}^2}=\pm \sqrt{25}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x+3=\pm 5\\ x+3-3=-3\pm 5[\text{Subtract}\text{3}\text{from}\text{both}\text{sides}]\\ x=-3+5\text{or}x=-3-5[\text{Seperate}\text{the}\text{solution}]\\ x=2\text{or}x=-8\end{array}$

The solutions are: − 8 and 2

Thus, the solutions are: − 8 and 2