Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 50


The real solutions to the equation $2y^{2}+7y+3=0$ are $y=-3$ and $y=-\frac{1}{2}$.

Work Step by Step

1. Factor $2y^{2}+7y+3$. The factors are $(y+3)(2y+1)$ because $y\times2y=2y^{2}$(the a value in a2+bx+c) and $3\times1=3$ the c value in the equation. When the factors are FOILED out, they equal the equation $2y^{2}+7y+3$. 2. Set the factors equal to zero: $y+3=0$ and $2y+1=0$ 3. Solve: $y=-3$ $y=-\frac{1}{2}$
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