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: 6

Answer

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

Work Step by Step

Step 1: Comparing $n^{2}-4n-1=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=-4$ and $c=-1$ 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: $x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(-1)}}{2(1)}$ Step 4: $x=\frac{4 \pm \sqrt {16+4}}{2}$ Step 5: $x=\frac{4 \pm \sqrt {20}}{2}$ Step 6: $x=\frac{4 \pm \sqrt {4\times5}}{2}$ Step 7: $x=\frac{4 \pm \sqrt {2^{2}\times5}}{2}$ Step 8: $x=\frac{4 \pm 2\sqrt {5}}{2}$ Step 9: $x=2\pm\sqrt 5$ Step 10: $x=2+\sqrt 5$ or $x=2-\sqrt 5$ Step 11: Therefore, the solution set is {$2-\sqrt 5,2+\sqrt 5$}.
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