Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4: 13

Answer

{$7 - 2\sqrt {17},7 + 2\sqrt {17}$}

Work Step by Step

Step 1: Comparing $n^{2}-14n-19=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=1$, $b=-14$ and $c=-19$ 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: $n=\frac{-(-14) \pm \sqrt {(-14)^{2}-4(1)(-19)}}{2(1)}$ Step 4: $n=\frac{14 \pm \sqrt {196+76}}{2}$ Step 5: $n=\frac{14 \pm \sqrt {272}}{2}$ Step 6: $n=\frac{14 \pm \sqrt {16\times17}}{2}$ Step 7: $n=\frac{14 \pm (\sqrt {16}\times\sqrt {17})}{2}$ Step 8: $n=\frac{14 \pm (4\times \sqrt {17})}{2}$ Step 9: $n=\frac{2(7 \pm 2\sqrt {17})}{2}$ Step 10: $n=7 - 2\sqrt {17}$ or $n=7 + 2\sqrt {17}$ Step 11: Therefore, the solution set is {$7 - 2\sqrt {17},7 + 2\sqrt {17}$}.
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