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 8.2 - The Quadratic Formula - Exercise Set - Page 607: 8


{$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}

Work Step by Step

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $4x^2+3x=2$ This can be re-written as: $4x^2+3x-2=0$ Thus, $x=\dfrac{-(3) \pm \sqrt{(3)^2-4(4)(-2)}}{2(4)}$ or, $x=\dfrac{-3 \pm \sqrt{41}}{8}$ or, $x=\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$ Our solution set is: {$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.