Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 148: 54


The solution is $x=16$

Work Step by Step

$\sqrt{x}-\sqrt{x-12}=2$ Take $\sqrt{x-12}$ to the right side: $\sqrt{x}=2+\sqrt{x-12}$ Square both sides: $(\sqrt{x})^{2}=(2+\sqrt{x-12})^{2}$ $x=4+4\sqrt{x-12}+x-12$ Leave the term with the square root alone on the right side of the equation: $x-4-x+12=4\sqrt{x-12}$ $8=4\sqrt{x-12}$ Take $4$ to divide the left side: $\dfrac{8}{4}=\sqrt{x-12}$ $2=\sqrt{x-12}$ Square both sides: $2^{2}=(\sqrt{x-12})^{2}$ $4=x-12$ Rearrange: $x-12=4$ Solve for $x$: $x=4+12$ $x=16$ The solution is $x=16$
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