## College Algebra (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 178: 6

#### Answer

The solutions are $x=-1$ and $x=\pm\dfrac{1}{3}$

#### Work Step by Step

$x+1=9x^{3}+9x^{2}$ Take all terms to the right side of the equation: $0=9x^{3}+9x^{2}-x-1$ Rearrange: $9x^{3}+9x^{2}-x-1=0$ Group the first two terms together and the last two terms together: $(9x^{3}+9x^{2})-(x+1)=0$ Take out common factor $9x^{2}$ from the first group: $9x^{2}(x+1)-(x+1)=0$ Factor out $(x+1)$: $(x+1)(9x^{2}-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+1=0$ $x=-1$ $9x^{2}-1=0$ $9x^{2}=1$ $x^{2}=\dfrac{1}{9}$ $\sqrt{x^{2}}=\pm\sqrt{\dfrac{1}{9}}$ $x=\pm\dfrac{1}{3}$ The solutions are $x=-1$ and $x=\pm\dfrac{1}{3}$

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