College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 54

Answer

$x=3$

Work Step by Step

We are given: $\sqrt{3\sqrt{2x+3}}=\sqrt{5x-6}$ We square both sides: $(\sqrt{3\sqrt{2x+3}})^{2}=(\sqrt{5x-6})^{2}$ $3\sqrt{2x+3}=5x-6$ And square both sides again: $(3\sqrt{2x+3})^{2}=(5x-6)^{2}$ $9\ (2x+3)=25x^{2}-60x+36$ And distribute: $18x+27=25x^{2}-60x+36$ $25x^{2}-78x+9=0$ Factor the trinomial: $(25x-3)(x-3)=0$ Use the zero-factor property by equating each factor to zero: $(25x-3)=0$ or $(x-3)=0$ $x=\displaystyle \frac{3}{25}$ or $x=3$ However, the solution $x=\frac{3}{25}$ does not work in the original equation: $\sqrt{3\sqrt{2(\frac{3}{25})+3}}=\sqrt{5*(\frac{3}{25})-6}$ $\sqrt{3\sqrt{\frac{6}{25}+3}}=\sqrt{\frac{3}{5}-6}$ $\sqrt{3\sqrt{\frac{6}{25}+\frac{75}{25}}}=\sqrt{\frac{3}{5}-\frac{30}{5}}$ $\sqrt{3\sqrt{\frac{81}{25}}}=\sqrt{-\frac{27}{5}}$ Since we got an untrue statement (the right side is imaginary), the solution $x=\frac{3}{25}$ does not work. Thus the only solution is $x=3$.
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