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: 60

Answer

$x=\left\{ \dfrac{4}{3},2 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt[3]{3x^2-9x+8}=\sqrt[3]{x} ,$ cube both sides of the equal sign. Then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property. $\bf{\text{Solution Details:}}$ Raising both sides to the third power and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 3x^2-9x+8=x \\\\ 3x^2+(-9x-x)+8=0 \\\\ 3x^2-10x+8=0 .\end{array} In the trinomial expression above, the value of $ac$ is $ 3(8) $ and the value of $b$ is $ -10 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -4,-6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2-4x-6x+8=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^2-4x)-(6x-8)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3x-4)-2(3x-4)=0 .\end{array} Factoring the $GCF= (3x-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-4)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 3x-4=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-4=0 \\\\ 3x=4 \\\\ x=\dfrac{4}{3} \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} Hence, $ x=\left\{ \dfrac{4}{3},2 \right\} .$
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