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: 115



Work Step by Step

The LHS is nonnegative, so the RHS = x is also nonnegative. (We will exclude any negative solution) $\sqrt{3x+18}=x \quad $... Square both sides $3x+18=x^{2}$ $ x^{2}-3x-18=0$ We factor this trinomial by finding two factors of $-18$ that add to $-3$; they are $+3$ and $-6$. $(x+3)(x-6)=0$ $ x=-3,\quad x=-6$ $ x=-3 $ is extraneous (x must be nonnegative), so we discard it. $ x=6$ is a valid solution Check: $\sqrt{3(6)+18}=6$ $ \sqrt{18+18}=6$ $\sqrt{36}=6$ The solution set is $\{6\}$.
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