Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.2 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 779: 70



Work Step by Step

$\dfrac{10}{z}=\dfrac{5}{z}-\dfrac{1}{3} \\ \dfrac{10}{z}=\dfrac{15-z}{3z} \\ 30z=z(15-z) \\ 30z=15z-z^2 \\ z^2+30z-15z=0 \\ z^2+15z=0 \\ z(z+15)=0 \\ $ But the solution is only $z=-15$, because if $z=0$ the original equation would be undefined.
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