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: 14

Answer

{$2 - i\sqrt 5,2 + i\sqrt 5$}

Work Step by Step

Step 1: Comparing $t^{2}-4t+9=0$ to the standard form of a quadratic equation, $at^{2}+bt+c=0$, we find: $a=1$, $b=-4$ and $c=9$ 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: $x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(9)}}{2(1)}$ Step 4: $x=\frac{4 \pm \sqrt {16-36}}{2}$ Step 5: $x=\frac{4 \pm \sqrt {-20}}{2}$ Step 6: $x=\frac{4 \pm \sqrt {-1\times20}}{2}$ Step 7: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {4\times5})}{2}$ Step 8: $x=\frac{4 \pm (i\times 2\sqrt 5)}{2}$ Step 9: $x=\frac{4 \pm i2\sqrt 5}{2}$ Step 10: $x=\frac{2(2 \pm i\sqrt 5)}{2}$ Step 11: $x=2 \pm i\sqrt 5$ Step 12: $x=2 - i\sqrt 5$ or $x=2 + i\sqrt 5$ Step 13: Therefore, the solution set is {$2 - i\sqrt 5,2 + i\sqrt 5$}.
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