Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 117



Work Step by Step

The LHS is nonnegative, so the RHS = $ x-3$ is also nonnegative. All solutions must satisfy $ x\geq 3\qquad (*)$ $\sqrt{x+3}=x-3 \quad $... Square both sides $ x+3=x^{2}-6x+9$ $ x^{2}-7x+6=0$ We factor this trinomial by finding two factors of $+6$ that add to $-7$; they are $-6$ and $-1$. $(x-6)(x-1)=0$ $ x=6,\quad x=1$ $ x=1 \;$ is extraneous (does not satisfy (*) ), so we discard it. $ x=6$ is a valid solution Check: $\sqrt{6+3}=6-3$ $ \sqrt{9}=3$ The solution set is $\{6\}$.
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