College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 58



Work Step by Step

We solve by factoring: $x+\sqrt{x}=6$ $x+\sqrt{x}-6=0$ $[\sqrt{x}+3][\sqrt{x}-2]=0$ $\sqrt{x}+3=0$ or $\sqrt{x}-2=0$ $\sqrt{x}=-3$ or $\sqrt{x}=2$ $x=(-3)^2$ or $x=2^2$ $x=9$ or $x=4$ However, $x=9$ does not work in the original equation, so we throw this solution out. $9+\sqrt{9}\ne 6$ Thus, the only solution is: $x=4$
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