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: 64

Answer

$x=\left\{ -8,2 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt[4]{x^2+6x}=2 ,$ raise both sides to the fourth power. 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. Checking of the solution is a must since both sides were raised to an even power. $\bf{\text{Solution Details:}}$ Raising both sides to the fourth power and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} x^2+6x=16 \\\\ x^2+6x-16=0 .\end{array} In the trinomial expression above, the value of $ac$ is $ 1(-16)=-16 $ and the value of $b$ is $ 6 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -2,8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-2x+8x-16=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} (x^2-2x)+(8x-16)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-2)+8(x-2)=0 .\end{array} Factoring the $GCF= (x-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-2)(x+8)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-2=0 \\\\\text{OR}\\\\ x+8=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-2=0 \\\\ x=2 \\\\\text{OR}\\\\ x+8=0 \\\\ x=-8 .\end{array} Upon checking, $ x=\left\{ -8,2 \right\} $ satisfy the original equation.
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