## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 70

#### Answer

The solutions are $\frac{1±\sqrt{55}}{6}$, and the solution set is {$\frac{1+\sqrt{55}}{6}, \frac{1-\sqrt{55}}{6}$}.

#### Work Step by Step

$\frac{x^2}{2}-\frac{x}{6}-\frac{3}{4}=0$ The least common multiple of $2$ and $6$ is $6$. Multiply the equation by the LCM. $\frac{x^2}{2}·6-\frac{x}{6}·6-\frac{3}{4}·6=0·6$ $3x^2-x-\frac{9}{2}=0$ Divide both sides by $3$. Thus, it becomes: $x^2-\frac{x}{3}-\frac{3}{2}=0$ Add $\frac{3}{2}$ to both sides: $x^2-\frac{x}{3}-\frac{3}{2}+\frac{3}{2}=0+\frac{3}{2}$ $x^2-\frac{x}{3}=\frac{3}{2}$ The coefficient of the x-term is $-\frac{1}{3}$. Half of $-\frac{1}{3}$ is $-\frac{1}{6}$, and $(-\frac{1}{6})^2$ is $\frac{1}{36}$. Add $\frac{1}{36}$ to both sides of the equation to complete the square. $x^2-\frac{x}{3}+\frac{1}{36}=\frac{3}{2}+\frac{1}{36}$ $(x-\frac{1}{6})^2 = \frac{55}{36}$ $x-\frac{1}{6} = ±\sqrt\frac{55}{36}$ $x = \frac{1}{6}±\sqrt\frac{55}{36}$ $x=\frac{1±\sqrt{55}}{6}$ The solutions are $\frac{1±\sqrt{55}}{6}$, and the solution set is {$\frac{1+\sqrt{55}}{6}, \frac{1-\sqrt{55}}{6}$}.

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