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: 55

Answer

{$-8.10,2.10$}

Work Step by Step

Step 1: Comparing $x^{2}+6x-17=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=6$, and $c=-17$ 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: $x=\frac{-(6) \pm \sqrt {(6)^{2}-4(1)(-17)}}{2(1)}$ Step 4: $x=\frac{-6 \pm \sqrt {36+68}}{2}$ Step 5: $x=\frac{-6 \pm \sqrt {104}}{2}$ Step 6: $x=\frac{-6 \pm \sqrt {4\times26}}{2}$ Step 7: $x=\frac{-6 \pm 2\sqrt {26}}{2}$ Step 8: $x=-3\pm\sqrt {26}$ Step 9: $x=-3\pm5.099$ Step 10: $x=-3+5.099$ or $x=-3-5.099$ Step 11: $x=2.099\approx2.10$ or $x=-8.099\approx-8.10$ Step 12: Therefore, the solution set is {$-8.10,2.10$}.
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