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 - Study Summary - Practice Exercises: 14

Answer

$x=3$

Work Step by Step

Squaring both sides of the given equation results to \begin{array}{l}\require{cancel} \sqrt{2x+3}=x \\\\ \left( \sqrt{2x+3} \right)^2=(x)^2 \\\\ 2x+3=x^2 \\\\ -x^2+2x+3=0 \\\\ -1(-x^2+2x+3)=-1(0) \\\\ x^2-2x-3=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2-2x-3=0 \\\\ (x-3)(x+1)=0 .\end{array} Equating each factor to $0$ (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} x-3=0 \\\\\text{ OR }\\\\ x+1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{ OR }\\\\ x+1=0 \\\\ x=-1 .\end{array} If $x=-1,$ then \begin{array}{l}\require{cancel} \sqrt{2x+3}=x \\\\= \sqrt{2(-1)+3}=-1 \\\\= \sqrt{-2+3}=-1 \\\\= \sqrt{1}=-1 \\\\= 1=-1 \text{ (FALSE)} .\end{array} Hence, only $ x=3 $ satisfies the original equation.
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