College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 100



Work Step by Step

Use the given information to create an equation to begin solving for the variable. Then square both sides to rationalize. Move all the terms to one side and factor. Then set each set of parentheses equal to zero to solve for the variable. The value $4$ makes the equation false, so $7$ is the solution. $\sqrt {x-3}=x-5$ $(\sqrt {x-3})^2=(x-5)^2$ $x-3=x^2-10x+25$ $0=x^2-11x+28$ $(x-4)(x-7)=0$ $x=4,7$ $x=7$
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