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: 118



Work Step by Step

For the root to be real, x must be $\geq-10.$ The LHS is nonnegative, so the RHS = $ x-2$ is also nonnegative. All solutions must satisfy $ x\geq 2\qquad (*)$ $\sqrt{x+10}=x-2 \quad $... Square both sides $ x+10=(x-2)^{2}$ $ x+10=x^{2}-4x+4$ $0=x^{2}-5x-6$ We factor this trinomial by finding two factors of $-6$ that add to $-5$; 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: $\sqrt{6+10}=6-2$ $ \sqrt{16}=4$ The solution set is $\{6\}$.
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