Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 637: 74



Work Step by Step

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $2x^2+x=1 5$ It can be written as: $2x^2+x-15=0$ Thus, $x=\dfrac{-(1) \pm \sqrt{(1)^2-4(2)(-15)}}{2(2)}$ or, $x=\dfrac{-1 \pm \sqrt{121}}{4}$ or, $x=\dfrac{-1+11}{4}=\dfrac{5}{2}$ and $x=\dfrac{-1-11}{4}=-3$ Our solution set is: {$-3,\dfrac{5}{2}$}
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