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 655: 85

Answer

$x= -6$, $x= 0$ or $x= 2$

Work Step by Step

Given \begin{equation} 2 x^3+8 x^2=24 x. \end{equation} This equation can be regarded as other form of equation and can be solved by factoring. \begin{equation} \begin{aligned} 2 x^3+8 x^2&=24 x\\ 2 x^3+8 x^2-24x&= 0\\ 2x(x^2+4x-12) & =0\\ 2x(x-2)(x+6)&= 0\\ \implies x&= 0\\ x&=2 \\ x&= -6. \end{aligned} \end{equation} The solution is $x= -6$, $x= 0$ or $x= 2$.
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