Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.1 - Complex Numbers - 8.1 Problem Set - Page 426: 85


Comparing the equation with the general form $ax^{2}+bx+c$, we get a=1, b= -4 and c=13. $b^{2}-4ac$ is less than 0. Therefore, the solutions are given by $x= \frac{-b±\sqrt {4ac-b^{2}}i}{2a}=\frac{4±\sqrt {4\times1\times13-16}i}{2}$ or $x=2+3i$ or $x=2-3i$. Hence proved.

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