Chapter 9.3, Problem 87S

To determine

To Solve: The equations by completing the square method.

Answer to Problem 87S

  $x=\frac{3\pm \sqrt{33}}{4}$

Explanation of Solution

Given: $2x^2-3x-3$

Given: $2x^2-3x-3$

The coefficient of $x^2$ must be equal to one

Thus

Solving

  $\begin{array}{l}2x^2-3x-3\\ x^2-\frac{3}{2}x-\frac{3}{2}=0\\ x^2-\frac{3}{2}x=\frac{3}{2}\\ x^2-\frac{3}{2}x+\frac{9}{16}=\frac{3}{2}+\frac{9}{16}\\ {\left(x-\frac{3}{4}\right)}^4=\frac{33}{16}\\ x-\frac{3}{4}=\sqrt{\frac{33}{16}}\\ x-\frac{3}{4}=\pm \frac{\sqrt{33}}{4}\\ x=\pm \frac{\sqrt{33}}{4}+\frac{3}{4}\\ x=\frac{3\pm \sqrt{33}}{4}\end{array}$

Hence, $x=\frac{3\pm \sqrt{33}}{4}$