Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Cumulative Review Exercises - Page 647: 4



Work Step by Step

$\sqrt {x-3} +5=x \implies (\sqrt {x-3})^2=(x-5)^2$ Re-arrange as: $ x^2 -10x +25 -(x-3) =0$ $\implies x^2-11x+28=0$ $\implies x(x-7)-4(x-7)=0$ or, $(x-4)(x-7)=0$ So, $ x=7,4$ Neglect the negative term of $ x=4$, because it does not belong to the domain of the variable; so we have $ x=7$
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