Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 12


$\displaystyle \frac{5}{2}$ and $-2$

Work Step by Step

Set the $RHS$ to 0 by subtracting 10 from both sides $2k^{2}-k=10\qquad.../-10$ $2k^{2}-k-10=0$ To factor the trinomial, we search for factors of $2\times(-10)=-20, $ whose sum is $-1:\qquad $ We find $-5$ and +4. $2k^{2}-k-10$=$2k^{2}+4k-5k-10$ ... and factor in pairs: $2k(k+2)-5(k+2)=(2k-5)(k+2)$ Our equation becomes $(2k-5)(k+2)=0$ By the zero product principle, $2k-5=0$ or $k+2=0$ $k=\displaystyle \frac{5}{2}$ or $k=-2$ The solutions are $\displaystyle \frac{5}{2}$ and $-2$.
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