## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 40

#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{6x+7}-9=x-7 ,$ use the properties of equality to isolate the radical and then square both sides. 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. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt{6x+7}=x-7+9 \\\\ \sqrt{6x+7}=x+2 .\end{array} 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} 6x+7=(x+2)^2 \\\\ 6x+7=[(x)^2+2(x)(2)+(2)^2] \\\\ 6x+7=x^2+4x+4 \\\\ -x^2+(6x-4x)+(7-4)=0 \\\\ -x^2+2x+3=0 \\\\ -1(-x^2+2x+3)=(0)(-1) \\\\ 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, $x=\left\{ -1,3 \right\}$ satisfy the original equation.

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