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: 89


When $y=4+2i$, $x=4-2i$ when $y=4-2i$, $x=4+2i$

Work Step by Step

Working from the first equation, $x+y=8$, we make $x$ the subject of the formula, $x=8-y$ We then sub this modified version of the first equation into the second equation , $xy=20$ to get $(8-y)y=20$. Expanding this equation and moving all terms to the left hand side, we get: $-y^{2}+8y-20=0$ The method used in the introduction then uses the quadratic method to solve for the equation. Therefore, we get: $y=\frac{-8\pm\sqrt{8^{2}-4(-1)(-20)}}{2(-1)}\\=4\pm\frac{\sqrt{64-80}}{-2}\\=4\pm\frac{\sqrt{-16}}{-2}\\=4\pm\frac{\sqrt{(16)(-1)}}{2}\\=4\pm2\sqrt{(-1)}\\=4\pm2i$ From this, we can deduce the corresponding values of $x$. When $y=4+2i$, $x=8-(4+2i)\\=4-2i$ when $y=4-2i$, $x=8-(4-2i)\\=4+2i$
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