College Algebra (6th Edition)

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

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.5: 134

Answer

$x = \frac{5}{2}$ OR $x = -3$

Work Step by Step

$$2x^2 + x = 15$$ $$2x^2 + x - 15 = 0$$ Since this cannot be factorized simply, we can use the Quadratic Formula to solve for $x$: $$x = \frac{-b \frac{+}{} \sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-1 \frac{+}{} \sqrt{1^2 - 4(2)(-15)}}{2(2)}$$ $$x = \frac{-1 \frac{+}{} \sqrt{121}}{4}$$ $$x = \frac{ -1 \frac{+}{} 11}{4}$$ $$x = \frac{10}{4} = \frac{5}{2}$$ $$OR$$ $$x = \frac{-12}{4} = -3$$
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