Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Test: Chapter 10 - Page 695: 20



Work Step by Step

Isolating the radical expression, the given expression is equivalent to \begin{array}{l}\require{cancel} x=\sqrt{3x+3}-1 \\\\ x+1=\sqrt{3x+3} .\end{array} Squaring both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} x+1=\sqrt{3x+3} \\\\ (x+1)^2=(\sqrt{3x+3})^2 \\\\ (x)^2+2(x)(1)+(1)^2=3x+3 \\\\ x^2+2x+1=3x+3 \\\\ x^2+2x-3x+1-3=0 \\\\ x^2-x-2=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to\begin{array}{l}\require{cancel} x^2-x-2=0 \\\\ (x-2)(x+1)=0 .\end{array} Equating each factor to zero (Zero Product Property) and solving for the variable, then the solutions are $ x=\{-1,2\} .$ Upon checking, both solutions, $ x=\{-1,2\} ,$ satisfy the original equation.
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