Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 97

Answer

$x=4$

Work Step by Step

$\sqrt{2x+1}+1=x$ Take the $1$ to substract to the right side of the equation: $\sqrt{2x+1}=x-1$ Let's square both sides of the equation: $(\sqrt{2x+1})^{2}=(x-1)^{2}$ $2x+1=x^{2}-2x+1$ Take all terms to the right side of the equation: $x^{2}-2x+1-2x-1=0$ Simplify: $x^{2}-4x=0$ Take out common factor $x$: $x(x-4)=0$ We get two solutions, which are: $x=0$ and $x=4$ Let's check our answers: With $x=0$: $\sqrt{2(0)+1}+1=0$ $2\ne0$ False With $x=4$ $\sqrt{2(4)+1}+1=4$ $4=4$ True So, our answer is $x=4$
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