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 - Mid-Chapter Check Point - Page 629: 12


$x = \frac{-3±\sqrt{41}}{4}$

Work Step by Step

$\frac{x^2}{3} + \frac{x}{2} = \frac{2}{3}$ Multiply the equation by the least common multiple $6$. $\frac{x^2}{3}⋅6 + \frac{x}{2}⋅6 = \frac{2}{3}⋅6$ $2x^2+3x = 4$ $2x^2+3x-4 = 0$ Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ $a=2$, $b=3$, $c=-4$ $x = \frac{-3±\sqrt{3^2-(4⋅2⋅-4)}}{(2⋅2)}$ $x = \frac{-3±\sqrt{9-(-32)}}{4}$ $x = \frac{-3±\sqrt{41}}{4}$
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.