Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - 8.4 Solving Radical Equations - 8.4 Exercises - Page 653: 24

Answer

$x= 32$

Work Step by Step

Given \begin{equation} 5 \sqrt{2 x}=40. \end{equation} Start by squaring both sides of the equation to eliminate the radical sign and solve for $x$. \begin{equation} \begin{aligned} 5 \sqrt{2 x}& = 40 \\ \left( 5 \sqrt{2 x} \right)^2&= \left( 40 \right)^2\\ 25\cdot 2 x& = 1600\\ x& =\frac{1600}{50}\\ &= 32. \end{aligned} \end{equation} Check. \begin{equation} \begin{aligned} 5 \sqrt{2 \cdot 32} & \stackrel{?}{=} 40 \\ 40& =40 \end{aligned} \end{equation} The solution is $x= 32$.
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