## 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 596: 116

#### Answer

$x = \frac{1}{3}$ $b^2 - 4ac = 0$

#### Work Step by Step

$9x^2 - 6x + 1 = 0$ is in the form $ax^2+bx+c$. To solve by factoring: Find two First terms whose product is $ax^2$: $(3x +$ __ $)$$(3x+$ __ $)$ Find two Last terms whose product is $c$: $(3x + 1 )(3x+ 1 )$ combination 1 $(3x - 1 )(3x- 1 )$ combination 2 Check for combination 1: $(3x + 1 )(3x+ 1 )$ $(3x + 1 )(3x+ 1 ) = (3x)(3x) +3x + 3x +1$ $(3x + 1 )(3x+ 1 ) = 9x^2 + 6x +1$ Check for combination 2: $(3x - 1 )(3x- 1 ) = (3x)(3x) - 3x-3x +1$ $(3x - 1 )(3x- 1 ) = 9x^2 - 6x +1$

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