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: 15



Work Step by Step

Step 1: Comparing $n^{2}-6n+13=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=1$, $b=-6$ and $c=13$ 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: $x=\frac{-(-6) \pm \sqrt {(-6)^{2}-4(1)(13)}}{2(1)}$ Step 4: $x=\frac{6 \pm \sqrt {36-52}}{2}$ Step 5: $x=\frac{6 \pm \sqrt {-16}}{2}$ Step 6: $x=\frac{6 \pm \sqrt {-1\times16}}{2}$ Step 7: $x=\frac{6 \pm (\sqrt {-1}\times\sqrt {4\times4})}{2}$ Step 8: $x=\frac{6 \pm (i\times 4)}{2}$ Step 9: $x=\frac{6 \pm 4i}{2}$ Step 10: $x=\frac{2(3 \pm 2i)}{2}$ Step 11: $x=3 \pm 2i$ Step 12: $x=3-2i$ or $x=3+2i$ Step 13: Therefore, the solution set is {$3-2i,3+2i$}.
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