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: 65

Answer

$x=-4$ or $x=1$

Work Step by Step

We solve by factoring: $x^{2}+3x+\sqrt{x^{2}+3x}=6$ $x^{2}+3x+\sqrt{x^{2}+3x}-6=0$ $[\sqrt{x^2+3x}+3][\sqrt{x^2+3x}-2]=0$ $\sqrt{x^2+3x}+3=0$ or $\sqrt{x^2+3x}-2=0$ $\sqrt{x^2+3x}=-3$ is not possible because a square root is never negative. So: $\sqrt{x^2+3x}=2$ $x^2+3x=(2)^2$ $x^2+3x=4$ $x^2+3x-4=0$ $(x+4)(x-1)=0$ $x=-4$ or $x=1$
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