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: 35


{$\frac{-2-\sqrt {5}}{2},\frac{-2+\sqrt {5}}{2}$}

Work Step by Step

Step 1: Comparing $4n^{2}+8n-1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=4$, $b=8$ and $c=-1$ Step 2: The quadratic formula is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b, and c in the formula: $n=\frac{-(8) \pm \sqrt {(8)^{2}-4(4)(-1)}}{2(4)}$ Step 4: $n=\frac{-8 \pm \sqrt {64+16}}{8}$ Step 5: $n=\frac{-8 \pm \sqrt {80}}{8}$ Step 6: $n=\frac{-8 \pm \sqrt {16\times5}}{8}$ Step 7: $n=\frac{-8 \pm 4\sqrt {5}}{8}$ Step 8: $n=\frac{4(-2 \pm \sqrt {5})}{8}$ Step 9: $n=\frac{(-2 \pm \sqrt {5})}{2}$ Step 10: $n=\frac{-2-\sqrt {5}}{2}$ or $n=\frac{-2+\sqrt {5}}{2}$ Step 11: Therefore, the solution set is {$\frac{-2-\sqrt {5}}{2},\frac{-2+\sqrt {5}}{2}$}.
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