Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 60


The solutions are $x=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$

Work Step by Step

$\dfrac{2}{3}x^{2}+\dfrac{1}{4}x=3$ Multiply the whole equation by $12$: $12\Big(\dfrac{2}{3}x^{2}+\dfrac{1}{4}x=3\Big)$ $8x^{2}+3x=36$ Take $36$ to the left side: $8x^{2}+3x-36=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=8$, $b=3$ and $c=-36$ Substitute the known values into the formula and evaluate: $x=\dfrac{-3\pm\sqrt{3^{2}-4(8)(-36)}}{2(8)}=\dfrac{-3\pm\sqrt{9+1152}}{16}=...$ $...=\dfrac{-3\pm\sqrt{1161}}{16}=\dfrac{-3\pm3\sqrt{129}}{16}=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$ The solutions are $x=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$
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