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: 116



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 $20-8x=x^{2}$ $ x^{2}+8x-20=0$ We factor this trinomial by finding two factors of $-20$ that add to $+8$; they are $+10$ and $-2$. $(x+10)(x-2)=0$ $ x=-10,\quad x=2$ $ x=-10 $ is extraneous (x must be nonnegative), so we discard it. $ x=2$ is a valid solution Check: $\sqrt{20-8(2)}=2$ $ \sqrt{20-16}=2$ $\sqrt{4}=2$ The solution set is $\{2\}$.
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