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 - Page 135: 59


$x=\left\{ \dfrac{2}{5},1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt[3]{5x^2-6x+2}-\sqrt[3]{x}=0 ,$ transpose the second term and then 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. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{5x^2-6x+2}=\sqrt[3]{x} .\end{array} Raising both sides to the third power and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 5x^2-6x+2=x \\\\ 5x^2+(-6x-x)+2=0 \\\\ 5x^2-7x+2=0 .\end{array} In the trinomial expression above, the value of $ac$ is $ 5(2)=10 $ and the value of $b$ is $ -7 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -2,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5x^2-2x-5x+2=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} (5x^2-2x)-(5x-2)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(5x-2)-(5x-2)=0 .\end{array} Factoring the $GCF= (5x-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5x-2)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 5x-2=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5x-2=0 \\\\ 5x=2 \\\\ x=\dfrac{2}{5} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $ x=\left\{ \dfrac{2}{5},1 \right\} .$
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