Chapter 9, Problem 13PT

To determine

Solve the equation by completing the square method.

Answer to Problem 13PT

The solutions are: − 2 and 3

Explanation of Solution

Given:

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

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: $x^2-x-6=0$

  $\begin{array}{l}{x}^2-x-6=0[\text{original}\text{equation}]\\ x^2-x-6+\frac{1}{4}=0+\frac{1}{4}[\text{Since}{(}^{\frac{1}{2}}=\frac{1}{4}\text{add}\text{to}\text{each}\text{side}]\\ x^2-x+\frac{1}{4}-6+6=\frac{1}{4}+6[\text{Add}\text{6}\text{to}\text{each}\text{side}]\\ $ {(x-\frac{1}{2})}^2=\frac{25}{4}$\sqrt{{(x-\frac{1}{2})}^2}=\pm \sqrt{\frac{25}{4}}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-\frac{1}{2}=\pm \frac{5}{2}\\ x-\frac{1}{2}+\frac{1}{2}=\frac{1}{2}\pm \frac{5}{2}[\text{Add}\frac{1}{2}\text{from}\text{both}\text{sides}]\\ x=\frac{1}{2}-\frac{5}{2}\text{or}x=\frac{1}{2}+\frac{5}{2}[\text{Seperate}\text{the}\text{solution}]\\ x=-2\text{or}x=3\end{array}$

The solutions are: − 2 and 3

Thus, the solutions are: − 2 and 3