Chapter 9.4, Problem 20PPS

To determine

Solve the equation by completing the square method.

Answer to Problem 20PPS

The solutions are: − 2.9 and 4.9

Explanation of Solution

Given:

The equation: $x^2-2x-14=0$

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: $x^2-2x-14=0$

  $\begin{array}{l}{x}^2-2x-14=0[\text{original}\text{equation}]\\ $ x^2-2x+1-14=0+1[\text{Since}{(\frac{2}{2})}^2=1\text{add}\text{to}\text{each}\text{side}]$x^2-2x+1-14+14=1+14[\text{add}\text{14}\text{to}\text{each}\text{side}]\\ $ {(x-1)}^2=15$\sqrt{{(x-1)}^2}=\pm \sqrt{15}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-1=\pm 3.9\\ x-1+1=1\pm 3.9[\text{add}\text{1}\text{to}\text{both}\text{sides}]\\ x=1-3.9\text{or}x=1+3.9[\text{Seperate}\text{the}\text{solution}]\\ x=-2.9\text{or}x=4.9\end{array}$

The solutions are: − 2.9 and 4.9

Thus, the solutions are: − 2.9 and 4.9