College Algebra 7th Edition

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

Chapter P, Prerequisites - Section P.8 - Solving Basic Equations - P.8 Exercises: 14

Answer

(a) yes (b) no

Work Step by Step

(a) Using $x=\displaystyle \frac{b}{2}$, we get: Left side: $(\displaystyle \frac{b}{2})^{2}-b(\frac{b}{2})+\frac{1}{4}b^{2}=\frac{b^{2}}{4}-\frac{b^{2}}{2}+\frac{b^{2}}{4}=0$ Right side: $0$ The sides match, so $x=b/2$ is a solution. (b) Using $x=\displaystyle \frac{1}{b}$, we get: Left side: $(\displaystyle \frac{1}{b})^{2}-b(\frac{1}{b})+\frac{1}{4}b^{2}=\frac{1}{b^{2}}-1+\frac{b^{2}}{4}$ Right side: $0$ The sides do not match, so $x=\displaystyle \frac{1}{b}$ is not a solution.
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