College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises: 68

Answer

$x=\left\{ -\dfrac{64}{3},4 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (3x^2+52x)^{1/4}=4 ,$ raise both sides to the exponent equal to $ 4 .$ 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 fourth power, the given equation becomes \begin{array}{l}\require{cancel} 3x^2+52x=256 \\\\ 3x^2+52x-256=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 3(-256)=-768 $ and the value of $b$ is $ 52 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -12,64 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2-12x+64x-256=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} (3x^2-12x)+(64x-256)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x(x-4)+64(x-4)=0 .\end{array} Factoring the $GCF= (x-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-4)(3x+64)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-4=0 \\\\\text{OR}\\\\ 3x+64=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-4=0 \\\\ x=4 \\\\\text{OR}\\\\ 3x+64=0 \\\\ 3x=-64 \\\\ x=-\dfrac{64}{3} .\end{array} Upon checking, $ x=\left\{ -\dfrac{64}{3},4 \right\} $ satisfy the original equation.
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