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: 12

Answer

{$-2 - 3\sqrt {2},-2 + 3\sqrt {2}$}

Work Step by Step

Step 1: Comparing $x^{2}+4x-14=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=1$, $b=4$ and $c=-14$ 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: $x=\frac{-(4) \pm \sqrt {(4)^{2}-4(1)(-14)}}{2(1)}$ Step 4: $x=\frac{-4 \pm \sqrt {16+56}}{2}$ Step 5: $x=\frac{-4 \pm \sqrt {72}}{2}$ Step 6: $x=\frac{-4 \pm \sqrt {36\times2}}{2}$ Step 7: $x=\frac{-4 \pm (\sqrt {36}\times\sqrt {2})}{2}$ Step 8: $x=\frac{-4 \pm (6\times \sqrt {2})}{2}$ Step 9: $x=\frac{2(-2 \pm 3\sqrt {2}}{2}$ Step 10: $x=-2 \pm 3\sqrt {2}$ Step 11: $x=-2 - 3\sqrt {2}$ or $x=-2 + 3\sqrt {2}$ Step 12: Therefore, the solution set is {$-2 - 3\sqrt {2},-2 + 3\sqrt {2}$}.
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