## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 73

$x=\left\{ \dfrac{1}{4},1 \right\}$$#### Work Step by Step$\bf{\text{Solution Outline:}}$To solve the given equation,$ (2x-1)^{2/3}=x^{1/3} ,$raise both sides to the third power to get rid of the denominator in the fractional exponent. Then express the resulting equation in the form$ax^2+bx+c=0,$and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation.$\bf{\text{Solution Details:}}$Raising both sides to the third power, the given equation becomes \begin{array}{l}\require{cancel} \left((2x-1)^{2/3}\right)^3=\left( x^{1/3} \right)^3 .\end{array} Using the Power Rule of the laws of exponents which is given by$\left( x^m \right)^p=x^{mp},$the expression above is equivalent to \begin{array}{l}\require{cancel} (2x-1)^{\frac{2}{3}\cdot3 }=x^{\frac{1}{3}\cdot3} \\\\ (2x-1)^2=x .\end{array} Using the square of a binomial which is given by$(a+b)^2=a^2+2ab+b^2$or by$(a-b)^2=a^2-2ab+b^2,$the expression above is equivalent to \begin{array}{l}\require{cancel} (2x)^2-2(2x)(1)+(1)^2=x \\\\ 4x^2-4x+1=x \\\\ 4x^2+(-4x-x)+1=0 \\\\ 4x^2-5x+1=0 .\end{array} Using factoring of trinomials, the value of$ac$in the trinomial expression above is$ 4(1)=4 $and the value of$b$is$ -5 .$The$2$numbers that have a product of$ac$and a sum of$b$are$\left\{ -1,-4 \right\}.$Using these$2$numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4x^2-x-4x+1=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (4x^2-x)-(4x-1)=0 .\end{array} Factoring the$GCF$in each group results to \begin{array}{l}\require{cancel} x(4x-1)-(4x-1)=0 .\end{array} Factoring the$GCF= (4x-1) $of the entire expression above results to \begin{array}{l}\require{cancel} (4x-1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 4x-1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4x-1=0 \\\\ 4x=1 \\\\ x=\dfrac{1}{4} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking,$ x=\left\{ \dfrac{1}{4},1 \right\}\$ satisfy the original equation.

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