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: 23



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: $x=\frac{6}{12}$ or $x=\frac{-8}{12}$ Step 9: $x=\frac{1}{2}$ or $x=\frac{-2}{3}$ Step 10: Therefore, the solution set is {$-\frac{2}{3},\frac{1}{2}$}.
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