College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 48


$x=\displaystyle \frac{10}{9}$ or $x=2$

Work Step by Step

We are given: $\sqrt{4x+1}-\sqrt{x-1}=2$ $\sqrt{4x+1}=2+\sqrt{x-1}$ We square both sides: $(\sqrt{4x+1})^{2}=(2+\sqrt{x-1})^{2}$ $4x+1=4+(x-1)+4\sqrt{x-1}$ $4x+1=3+x+4\sqrt{x-1}$ $3x-2=4\sqrt{x-1}$ We square both sides again: $(3x-2)^{2}=(4\sqrt{x-1})^{2}$ $9x^{2}-12x+4=16(x-1)$ And distribute: $9x^{2}-12x+4=16x-16$ $9x^{2}-28x+20=0$ And factor: $(9x-10)(x-2)=0$ Use the zero-factor property by equating each factor to zero: $(9x-10)=0$ or $(x-2)=0$ $x=\displaystyle \frac{10}{9}$ or $x=2$
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