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: 37

Answer

$x=-1$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt{3x+7}=3x+5 ,$ square both sides and 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. Finally, it is a must to do checking of the solution. $\bf{\text{Solution Details:}}$ Squaring both sides and then using the special product on squaring binomials which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3x+7=(3x+5)^2 \\\\ 3x+7=[(3x)^2+2(3x)(5)+(5)^2] \\\\ 3x+7=9x^2+30x+25 \\\\ -9x^2+(3x-30x)+(7-25)=0 \\\\ -9x^2-27x-18=0 \\\\ \dfrac{-9x^2-27x-18}{-9}=\dfrac{0}{-9} \\\\ x^2+3x+2=0 .\end{array} In the trinomial expression above, the value of $ac$ is $ 1(2)=2 $ and the value of $b$ is $ 3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 1,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2+x+2x+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} (x^2+x)+(2x+2)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x+1)+2(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+2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x+1=0 \\\\\text{OR}\\\\ x+2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x+2=0 \\\\ x=-2 .\end{array} Upon checking, only $ x=-1 $ satisfies the original equation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.