College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises - Page 164: 74

Answer

$x=-1$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \sqrt{2x+3}=x+2 ,$ square both sides and then express in the form $ax^2+bx+c=0.$ Use concepts of solving quadratic equations to find the values of $x$. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the given equation results to \begin{array}{l}\require{cancel} 2x+3=(x+2)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2x+3=(x)^2+2(x)(2)+(2)^2 \\\\ 2x+3=x^2+4x+4 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(2x-4x)+(3-4)=0 \\\\ -x^2-2x-1=0 \\\\ -1(-x^2-2x-1)=-1(0) \\\\ x^2+2x+1=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (x+1)(x+1)=0 \\\\ (x+1)^2=0 .\end{array} Taking the square root of both sides results to \begin{array}{l}\require{cancel} x+1=\pm\sqrt{0} \\\\ x+1=0 \\\\ x=-1 .\end{array} Upon checking, $ 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.