College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding: 58

Answer

$x=3$ or $x=-\displaystyle \frac{3}{2}$

Work Step by Step

We solve: $\frac{2}{3}x^2-x-3=0$ We multiply by $3$: $3(\displaystyle \frac{2}{3}x^{2}-x-3)=3*0$ $2x^{2}-3x-9=0$ We solve using the quadratic formula ($a=2,\ b=-3,\ c=-9$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\displaystyle \frac{-(-3)\pm\sqrt{(-3)^{2}-4*2*-9}}{2*2}$ $x=\displaystyle \frac{3\pm\sqrt{9+72}}{4}$ $x=\frac{3\pm\sqrt{81}}{4}$ $x=\frac{3\pm 9}{4}$ $x=\displaystyle \frac{3+9}{4}$ or $x=\displaystyle \frac{3-9}{4}$ $x=3$ or $x=-\displaystyle \frac{3}{2}$
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