College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 53


The solutions are $x=\dfrac{1}{4}$ and $x=1$

Work Step by Step

$2x-3x^{1/2}+1=0$ Let $u$ be equal to $x^{1/2}$ If $u=x^{1/2}$, then $u^{2}=x$ Rewrite the original equation using the new variable $u$: $2u^{2}-3u+1=0$ Solve by factoring: $(2u-1)(u-1)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $2u-1=0$ $2u=1$ $u=\dfrac{1}{2}$ $u-1=0$ $u=1$ Substitute $u$ back to $x^{1/2}$ and solve for $x$: $x^{1/2}=\dfrac{1}{2}$ $(x^{1/2})^{2}=\Big(\dfrac{1}{2}\Big)^{2}$ $x=\dfrac{1}{4}$ $x^{1/2}=1$ $(x^{1/2})^{2}=1^{2}$ $x=1$ The solutions are $x=\dfrac{1}{4}$ and $x=1$
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