Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises - Page 512: 90

Answer

$x=\pm\dfrac{5\sqrt{a}}{3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 9x^2-25a=0 ,$ isolate first the squared variable. Then take the square root of both sides (Square Root Property) and simplify the resulting radical. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 9x^2=25a \\\\ x^2=\dfrac{25a}{9} .\end{array} Taking the square root of both sides and then simplifying the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{\dfrac{25a}{9}} \\\\ x=\pm\sqrt{\dfrac{25}{9}\cdot a} \\\\ x=\pm\sqrt{\left( \dfrac{5}{3} \right)^2\cdot a} \\\\ x=\pm\dfrac{5}{3}\sqrt{a} \\\\ x=\pm\dfrac{5\sqrt{a}}{3} .\end{array}
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