## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 501: 28

#### Answer

The solutions are $x=-\frac{27}{8}$ and $x=-\frac{125}{8}$.

#### Work Step by Step

Add $15$ to both sides of the equation: $4x^{2/3}+16x^{1/3}+15=0 \\4(x^{1/3})^2+16x^{1/3}+15=0$ Let $u=x^{1/3}$. Rewrite the equation above using $u$ to obtain: $4u^2+16u+15=0$ Factor the trinomial to obtain: $(2u+3)(2u+5)=0$ Factor each factor with a variable to zero then solve each equation to obtain: \begin{array}{ccc} &2u+3=0 &\text{or} &2u+5=0 \\&2u=-3 &\text{or} &2u=-5 \\&u=-\frac{3}{2} &\text{or} &u=-\frac{5}{2} \end{array} Since $u=x^{1/3}$, then: \begin{array}{ccc} &u=-\frac{3}{2} &\text{or} &u=-\frac{5}{2} \\&x^{1/3}=-\frac{3}{2} &\text{or} &x^{1//3}=-\frac{5}{2} \\&(x^{1/3})^3=(-\frac{3}{2})^3 &\text{or} &(x^{1//3})^3=(-\frac{5}{2})^3 \\&x=-\frac{27}{8} &\text{or} &x=-\frac{125}{8} \end{array} Therefore, the solutions are $x=-\frac{27}{8}$ and $x=-\frac{125}{8}$.

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