Elementary Algebra

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

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 467: 2


{$-4-\sqrt {13},-4+\sqrt {13}$}

Work Step by Step

Step 1: Comparing $x^{2}+8x+3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=1$, $b=8$ and $c=3$ 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: $x=\frac{-(8) \pm \sqrt {(8)^{2}-4(1)(3)}}{2(1)}$ Step 4: $x=\frac{-8 \pm \sqrt {64-12}}{2}$ Step 5: $x=\frac{-8 \pm \sqrt {52}}{2}$ Step 6: $x=\frac{-8 \pm \sqrt {4\times13}}{2}$ Step 7: $x=\frac{-8 \pm (\sqrt {4}\times\sqrt {13})}{2}$ Step 8: $x=\frac{-8 \pm (2\times \sqrt {13})}{2}$ Step 9: $x=\frac{2(-4 \pm \sqrt {13})}{2}$ Step 10: $x=-4 \pm \sqrt {13}$ Step 11: $x=-4 - \sqrt {13}$ or $x=-4 + \sqrt {13}$ Step 12: Therefore, the solution set is {$-4-\sqrt {13},-4+\sqrt {13}$}.
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