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: 86


The solution set is $\left\{-\frac{2}{3}, \frac{1}{2}\right\}$.

Work Step by Step

Subtract $2$ on both sides of the equation to obtain: $2-2 = y+6y^2-2 \\0 = 6y^2+y-2$ RECALL: A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ The given quadratic equation has: $a=6 \\b=1 \\c=-2$ Substitute these values into the quadratic formula to obtain: $z=\dfrac{-1 \pm \sqrt{1^2-4(6)(-2)}}{2(6)} \\z=\dfrac{-1 \pm \sqrt{1-(-48)}}{12} \\z=\dfrac{-1\pm \sqrt{1+48}}{12} \\x =\dfrac{-1\pm\sqrt{49}}{12} \\x=\dfrac{-1\pm\sqrt{7^2}}{12} \\x=\dfrac{-1\pm 7}{12}$ Split the solutions to obtain: $x_1 = \dfrac{-1+7}{12} = \dfrac{6}{12} = \dfrac{1}{2} \\x_2=\dfrac{-1-7}{12}=\dfrac{-8}{12}=-\dfrac{2}{3}$ Therefore, the solution set is $\left\{-\frac{2}{3}, \frac{1}{2}\right\}$.
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