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: 47

Answer

{$-14,-9$}

Work Step by Step

Step 1: Comparing $n^{2}+23n+126=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=23$, and $c=126$ 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, we obtain: $x=\frac{-(23) \pm \sqrt {(23)^{2}-4(1)(126)}}{2(1)}$ Step 4: $x=\frac{-23 \pm \sqrt {529-504}}{2}$ Step 5: $x=\frac{-23 \pm \sqrt {25}}{2}$ Step 6: $x=\frac{-23 \pm 5}{2}$ Step 7: $x=\frac{-23+5}{2}$ or $x=\frac{-23-5}{2}$ Step 8: $x=\frac{-18}{2}$ or $x=\frac{-28}{2}$ Step 9: $x=-9$ or $x=-14$ Step 10: Therefore, the solution set is {$-14,-9$}.
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