Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 501: 37


The solutions are $27$ and $125$.

Work Step by Step

The given equation can be written as: $(x^{1/3})^2-8x^{1/3}+15=0$ Let $u=x^{1/3}$. Rewrite the equation above using $u$ to obtain: $u^2-8u+15=0$ Factor the trinomial to obtain: $(u-5)(u-3)=0$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&u-5=0 &\text{or} &u-3=0 \\&u=5 &\text{or} &u=3 \end{array} Since $u=x^{1/3}$, then: \begin{array}{ccc} \\&u=5 &\text{or} &u=3 \\&x^{1/3}=5 &\text{or} &x^{1/3}=3 \\&(x^{1/3})^3=5^3 &\text{or} &(x^{1/3})^3=3^3 \\&x=125 &\text{or} &x=27 \end{array} Therefore, the solutions are $27$ and $125$.
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