## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 53

#### Answer

$x=-\displaystyle \frac{2}{9}$ or $x=2$

#### Work Step by Step

We are given: $\sqrt{2\sqrt{7x+2}}=\sqrt{3x+2}$ We square both sides: $(\sqrt{2\sqrt{7x+2}})^{2}=(\sqrt{3x+2})^{2}$ $2\sqrt{7x+2}=3x+2$ And square both sides again: $(2\sqrt{7x+2})^{2}=(3x+2)^{2}$ $(2\sqrt{7x+2})(2\sqrt{7x+2})=(3x+2)(3x+2)$ $4\ (7x+2)=9x^{2}+12x+4$ And distribute: $28x+8=9x^{2}+12x+4$ $9x^{2}-16x-4=0$ And factor: $(9x+2)(x-2)=0$ Use the zero-factor property by equating each factor to zero: $(9x+2)=0$ or $(x-2)=0$ $x=-\displaystyle \frac{2}{9}$ or $x=2$

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