College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 40

Answer

$x=-1$ or $x=2$

Work Step by Step

$\sqrt{3+x}=\sqrt{x^{2}+1}$ $(\sqrt{3+x})^{2}=(\sqrt{x^{2}+1})^{2}$ $3+x=x^{2}+1$ $0=x^{2}+1-x-3$ $x^{2}-x-2=0$ $(x+1)(x-2)=0$ $x+1=0$ or $x-2=0$ $x=-1$ or $x=2$ We need to confirm the solutions. First we confirm $x=-1$ Left side=$\sqrt{3+-1}=\sqrt{2}$ Right side=$\sqrt{(-1)^{2}+1}=\sqrt{1+1}=\sqrt{2}$ Which is true. Next we confirm $x=2$ Left side=$\sqrt{3+2}=\sqrt{5}$ Right side=$\sqrt{2^{2}+1}=\sqrt{5}$ Which is true.
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