Elementary Algebra

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

Chapter 10 - Quadratic Equations - Chapter 10 Test: 9

Answer

{$-1 - \sqrt {10},-1 +\sqrt {10}$}

Work Step by Step

Step 1: Comparing $n^{2}+2n-9=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=1$, $b=2$ and $c=-9$ Step 2: The quadratic formula to be used is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $n=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(-9)}}{2(1)}$ Step 4: $n=\frac{-2 \pm \sqrt {4+36}}{2}$ Step 5: $n=\frac{-2 \pm \sqrt {40}}{2}$ Step 6: $n=\frac{-2 \pm \sqrt {4\times10}}{2}$ Step 7: $n=\frac{-2 \pm (\sqrt {4}\times\sqrt {10})}{2}$ Step 8: $n=\frac{-2 \pm (2\times \sqrt {10})}{2}$ Step 9: $n=\frac{2(-1 \pm 1\sqrt {10})}{2}$ Step 10: $n=-1 - 1\sqrt {10}$ or $n=-1 + 1\sqrt {10}$ Step 11: Therefore, the solution set is {$-1 - \sqrt {10},-1 +\sqrt {10}$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.