Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 10 - Radical Functions and Equations - 10.3 - Solving Radical Equations - Exercises - Page 564: 32

Answer

The solution is $v=6$.

Work Step by Step

The given equation is $\Rightarrow \sqrt{2v-5}=\sqrt{\frac{v}{3}+5}$ Square each side of the equation. $\Rightarrow (\sqrt{2v-5})^2=(\sqrt{\frac{v}{3}+5})^2$ Simplify. $\Rightarrow 2v-5=\frac{v}{3}+5$ Add $5-\frac{v}{3}$ to each side. $\Rightarrow 2v-5+5-\frac{v}{3}=\frac{v}{3}+5+5-\frac{v}{3}$ Simplify. $\Rightarrow \frac{6v-v}{3}=10$ $\Rightarrow \frac{5v}{3}=10$ Multiply each side by $\frac{3}{5}$. $\Rightarrow \frac{3}{5}\times \frac{5v}{3}=\frac{3}{5}\times 10$ Simplify. $\Rightarrow v=6$ Check $v=6$. $\Rightarrow \sqrt{2v-5}=\sqrt{\frac{v}{3}+5}$ $\Rightarrow \sqrt{2(6)-5}=\sqrt{\frac{6}{3}+5}$ $\Rightarrow \sqrt{12-5}=\sqrt{2+5}$ $\Rightarrow \sqrt{7}=\sqrt{7}$ True. Hence, the solution is $v=6$.
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