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: 66


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

Work Step by Step

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