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 107: 124



Work Step by Step

Rewrite the equation so the root is isolated on one side (with a + sign) Add $6$ to both sides. $\sqrt{2x+15}=x+6$ For the root to be real, x must be such that $ x\geq-15/2.$ The RHS is nonnegative, so the LHS = $ x+6 \;$ is also nonnegative. All solutions must satisfy $ x\geq-6\qquad (*)$ Square both sides. $2x+15=x^{2}+12x+36$ $0=x^{2}+10x+21$ We factor this trinomial by finding two factors of $+21$ that add to $+10$; they are $+3$ and $+7$. $0=(x+3)(x+7)$ $ x=-7 \;$ does not satisfy (*). We discard it. $ x=-3 \;$ is a valid solution (satisfies (*) ) Check: $\sqrt{2(-3)+15}-6=-3$ $\sqrt{9}-6=-3$ $3-6=-3$ The solution set is $\{-3\}$.
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