Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3 - Page 453: 26

Answer

{$-\frac{1}{2},\frac{2}{3}$}

Work Step by Step

Step 1: Comparing $6x^{2}-x-2=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=6$, $b=-1$, and $c=-2$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b, and c in the formula, we obtain: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(6)(-2)}}{2(6)}$ Step 4: $x=\frac{1 \pm \sqrt {1+48}}{12}$ Step 5: $x=\frac{1 \pm \sqrt {49}}{12}$ Step 6: $x=\frac{1 \pm 7}{12}$ Step 7: $x=\frac{1 + 7}{12}$ or $x=\frac{1 - 7}{12}$ Step 8: $x=\frac{8}{12}$ or $x=\frac{-6}{12}$ Step 9: $x=\frac{2}{3}$ or $x=-\frac{1}{2}$ Step 10: Therefore, the solution set is {$-\frac{1}{2},\frac{2}{3}$}.
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