Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 501: 5

Answer

$x=\{1,4\}$

Work Step by Step

Squaring both sides of the given equation, $ \sqrt{9x}=x+2 ,$ then, \begin{array}{l}\require{cancel} \left( \sqrt{9x} \right)^2=\left( x+2 \right)^2 \\\\ 9x=(x)^2+2(x)(2)+(2)^2 \\\\ 9x=x^2+4x+4 \\\\ -x^2+9x-4x-4=0 \\\\ -x^2+5x-4=0 .\end{array} Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $ -x^2+5x-4=0 ,$ are \begin{array}{l}\require{cancel} \dfrac{-(5)\pm\sqrt{(5)^2-4(-1)(-4)}}{2(-1)} \\\\= \dfrac{-5\pm\sqrt{25-16}}{-2} \\\\= \dfrac{-5\pm\sqrt{9}}{-2} \\\\= \dfrac{-5\pm\sqrt{(3)^2}}{-2} \\\\= \dfrac{-5\pm3}{-2} \\\\= \dfrac{-5-3}{-2} \text{ OR } \dfrac{-5+3}{-2} \\\\= \dfrac{-8}{-2} \text{ OR } \dfrac{-2}{-2} \\\\= 4 \text{ OR } 1 .\end{array} Upon checking, both solutions satisfy the original equation. Hence, $ x=\{1,4\} .$
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