College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-7): 3

Answer

$x = 9$

Work Step by Step

$$x - 5 = \sqrt {x + 7}$$ $(x - 5)^{2} = (\sqrt {x + 7})^{2}$ $(x^{2} - 10x + 25) = (x + 7)$ $x^{2} - 10x - x + 25 - 7 = 0$ $x^{2} - 11x + 18 = 0$; which can be easily factorized, since two factors of $18$ that, when added, yield $-11$, are ($-9\times-2$): $(x - 9)(x - 2) = 0$ $x = 9$ or $x = 2$ By substituting in the original equation to corroborate, we find that only $x=9$ satisfies the conditions since $x=2$ would produce $\sqrt{9} = -3$.
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