College Algebra (11th Edition)

Published by Pearson

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 63

Answer

$x=\left\{ -3,1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt[4]{x^2+2x}=\sqrt[4]{3} ,$ 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+2x=3 \\\\ x^2+2x-3=0 .\end{array} In the trinomial expression above, the value of $ac$ is $1(-3)=-3$ and the value of $b$ is $2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-x+3x-3=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-x)+(3x-3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-1)+3(x-1)=0 .\end{array} Factoring the $GCF= (x-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-1)(x+3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ x+3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ x+3=0 \\\\ x=-3 .\end{array} Upon checking, $x=\left\{ -3,1 \right\}$ satisfy the original equation.

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