Precalculus (6th Edition)

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

Chapter 1 - Equations and Inequalities - Summary Exercises on Solving Equations - Exercises: 19

Answer

$x=3$

Work Step by Step

Cube both sides of the equation to obtain: $$((14-2x)^{2/3})^3=4^3 \\(14-2x)^2=64$$ Take the square root of both sides of the equation to obtain: $$\sqrt{(14-2x)^2}=\pm\sqrt{64} \\14-2x=\pm8$$ Subtract $14$ to both sides of the equation to obtain: $$14-2x-14 =\pm8 - 14 \\-2x = -14\pm8$$ Divide $-2$ to both sides of the equation to obtain: $$\dfrac{-2x}{-2}=\dfrac{-14\pm8}{-2} \\x=7 \mp 4 \\x_1=7-4=3 \\x_2=7+4=11$$ However, when $x=11$, the given equation becomes $$(14-2\cdot11)^{2/3} = 4 \\(14-22)^{2/3}=4 \\(-11)^{2/3}\ne4$$ This means that $11$ is an extraneous solution. Thus, the solution to the given equation is $3$.
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