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: 58


The solutions are $x=-\dfrac{1}{4}\pm\dfrac{\sqrt{39}}{12}i$

Work Step by Step

$-6x^{2}=3x+2$ Take all terms to the left side of the equation: $-6x^{2}-3x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=-6$, $b=-3$ and $c=-2$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(-6)(-2)}}{2(-6)}=\dfrac{3\pm\sqrt{9-48}}{-12}=...$ $...=\dfrac{3\pm\sqrt{-39}}{-12}=\dfrac{3\pm\sqrt{39}i}{-12}=-\dfrac{3}{12}\pm\dfrac{\sqrt{39}}{12}i=...$ $...=-\dfrac{1}{4}\pm\dfrac{\sqrt{39}}{12}i$ The solutions are $x=-\dfrac{1}{4}\pm\dfrac{\sqrt{39}}{12}i$
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