## 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: 69

#### Answer

The solutions are $\frac{-1±\sqrt{19}}{6}$, and the solution set is {$\frac{-1+\sqrt{19}}{6}, \frac{-1-\sqrt{19}}{6}$}.

#### Work Step by Step

$\frac{x^2}{3}+\frac{x}{9}-\frac{1}{6}=0$ The least common multiple of $3$ and $9$ is $9$. Multiply the equation by the LCM. $\frac{x^2}{3}·9+\frac{x}{9}·9-\frac{1}{6}·9=0·9$ $3x^2+x-\frac{3}{2}=0$ Divide both sides by $3$. Thus, it becomes: $x^2+\frac{x}{3}-\frac{1}{2}=0$ Add $\frac{1}{2}$ to both sides. $x^2+\frac{x}{3}-\frac{1}{2}+\frac{1}{2}=0+\frac{1}{2}$ $x^2+\frac{x}{3}=\frac{1}{2}$ 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. $x^2+\frac{x}{3}+\frac{1}{36}=\frac{1}{2}+\frac{1}{36}$ $(x+\frac{1}{6})^2 = \frac{19}{36}$ $x+\frac{1}{6}=\sqrt\frac{19}{36}$ or $x+\frac{1}{6}=-\sqrt\frac{19}{36}$ $x=-\frac{1}{6}+\sqrt\frac{19}{36}$ or $x=-\frac{1}{6}-\sqrt\frac{19}{36}$ $x=-\frac{1}{6}+\frac{\sqrt19}{6}$ or $x=x=-\frac{1}{6}-\frac{\sqrt19}{6}$ The solutions are $\frac{-1±\sqrt{19}}{6}$, and the solution set is {$\frac{-1+\sqrt{19}}{6}, \frac{-1-\sqrt{19}}{6}$}.

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