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: 5


{$1-\sqrt 6,1+\sqrt 6$}

Work Step by Step

Step 1: Comparing $n^{2}-2n-5=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=-2$ and $c=-5$ 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: $x=\frac{-(-2) \pm \sqrt {(-2)^{2}-4(1)(-5)}}{2(1)}$ Step 4: $x=\frac{2 \pm \sqrt {4+20}}{2}$ Step 5: $x=\frac{2 \pm \sqrt {24}}{2}$ Step 6: $x=\frac{2 \pm \sqrt {4\times6}}{2}$ Step 7: $x=\frac{2 \pm \sqrt {2^{2}\times6}}{2}$ Step 8: $x=\frac{2 \pm 2\sqrt {6}}{2}$ Step 9: $x=1\pm\sqrt 6$ Step 10: $x=1+\sqrt 6$ or $x=1-\sqrt 6$ Step 11: Therefore, the solution set is {$1-\sqrt 6,1+\sqrt 6$}.
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