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: 3



Work Step by Step

Step 1: Comparing $x^{2}+5x-36=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=5$ and $c=-36$ 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: $x=\frac{-(5) \pm \sqrt {(5)^{2}-4(1)(-36)}}{2(1)}$ Step 4: $x=\frac{-5 \pm \sqrt {25+144}}{2}$ Step 5: $x=\frac{-5 \pm \sqrt {169}}{2}$ Step 6: $x=\frac{-5 \pm 13}{2}$ Step 7: $x=\frac{-5+13}{2}$ or $x=\frac{-5-13}{2}$ Step 8: $x=\frac{8}{2}$ or $x=\frac{-18}{2}$ Step 9: $x=4$ or $x=-9$ Step 10: Therefore, the solution set is {$-9,4$}.
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