College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.8 - Solving Basic Equations - P.8 Exercises: 74

Answer

$x=-2+\dfrac{\sqrt[5]{8}}{2}$

Work Step by Step

Divide 4 to both sides of the equation to obtain: $(x+2)^5=\frac{1}{4}$ Take the fifth root of both sides: $x+2=\sqrt[5]{\frac{1}{4}}$ Subtract 2 to both sides of the equation to obtain: $x=-2 + \sqrt[5]{\frac{1}{4}}$ Rationalize the denominator by multiplying 8 to both the numerator and the denominator inside the radical sign to obtain: $x = -2 +\sqrt[5]{\frac{1}{4} \cdot \frac{8}{8}} \\x=-2 + \sqrt[5]{\frac{8}{32}} \\x=-2+\sqrt[5]{\frac{8}{2^5}} \\x=-2+\dfrac{\sqrt[5]{8}}{2}$ Thus, the solution to the given equation is $x=-2+\dfrac{\sqrt[5]{8}}{2}$.
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