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: 119



Work Step by Step

For the root to be real, x must be such that $ x\geq-13/2.$ The LHS is nonnegative, so the RHS = $ x+7 \;$ is also nonnegative. All solutions must satisfy $ x\geq-13/2\qquad (*)$ $2x+13=(x+7)^{2}$ $2x+13=x^{2}+14x+49$ $ x^{2}+12x+36=0 \quad $... recognize a perfect square $(x+6)^{2}=0$ $ x=-6 \;$ is a valid solution (satisfies (*) ) Check: $\sqrt{2(-6)+13}=-6+7$ $\sqrt{-12+13}=1$ $\sqrt{1}=1$ The solution set is $\{-6\}$.
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