Elementary Algebra

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

Chapter 11 - Additional Topics - 11.5 - Quadratic Equations: Complex Solutions - Problem Set 11.5 - Page 496: 18



Work Step by Step

Step 1: Comparing $x^{2}+2x+5=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=1$, $b=2$ and $c=5$ 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: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(5)}}{2(1)}$ Step 4: $x=\frac{-2 \pm \sqrt {4-20}}{2}$ Step 5: $x=\frac{-2 \pm \sqrt {-16}}{2}$ Step 6: $x=\frac{-2 \pm \sqrt {-1\times16}}{2}$ Step 7: $x=\frac{-2 \pm (\sqrt {-1}\times\sqrt {16})}{2}$ Step 8: $x=\frac{-2 \pm (i\times 4)}{2}$ Step 9: $x=\frac{-2 \pm 4i}{2}$ Step 10: $x=\frac{2(-1 \pm 2i)}{2}$ Step 11: $x=-1 \pm 2i$ Step 12: $x=-1-2i$ or $x=-1+2i$ Step 13: Therefore, the solution set is {$-1-2i,-1+2i$}.
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