College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.4: 26

Answer

$\dfrac{-6i}{3+2i}=-\dfrac{12}{13}-\dfrac{18}{13}i$

Work Step by Step

$\dfrac{-6i}{3+2i}$ Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator: $\dfrac{-6i}{3+2i}=\dfrac{-6i}{3+2i}\cdot\dfrac{3-2i}{3-2i}=\dfrac{-6i(3-2i)}{3^{2}-(2i)^{2}}=\dfrac{-18i+12i^{2}}{9-4i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{-18i+12(-1)}{9-4(-1)}=\dfrac{-18i-12}{9+4}=\dfrac{-12-18i}{13}=...$ $...=-\dfrac{12}{13}-\dfrac{18}{13}i$
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