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: 54

Answer

{$-10.44,1.44$}

Work Step by Step

Step 1: Comparing $x^{2}+9x-15=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=9$, and $c=-15$ 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: $x=\frac{-(9) \pm \sqrt {(9)^{2}-4(1)(-15)}}{2(1)}$ Step 4: $x=\frac{-9 \pm \sqrt {81+60}}{2}$ Step 5: $x=\frac{-9 \pm \sqrt {141}}{2}$ Step 6: $x=\frac{-9 \pm 11.874}{2}$ Step 7: $x=\frac{-9+11.874}{2}$ or $x=\frac{-9-11.874}{2}$ Step 8: $x=\frac{2.874}{2}$ or $x=\frac{-20.874}{2}$ Step 9: $x=1.437\approx1.44$ or $x=-10.437\approx-10.44$ Step 10: Therefore, the solution set is {$-10.44,1.44$}.
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