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 - Page 143: 54



Work Step by Step

$\dfrac{x^{2}+11}{3-x}$ for $x=4i$ Substitute $x$ by $4i$: $\dfrac{(4i)^{2}+11}{3-4i}=\dfrac{16i^{2}+11}{3-4i}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{16(-1)+11}{3-4i}=\dfrac{-16+11}{3-4i}=\dfrac{-5}{3-4i}=...$ Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator: $...=\dfrac{-5}{3-4i}\cdot\dfrac{3+4i}{3+4i}=\dfrac{-5(3+4i)}{3^{2}-(4i)^{2}}=\dfrac{-15-20i}{9-16i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{-15-20i}{9-16(-1)}=\dfrac{-15-20i}{9+16}=\dfrac{-15-20i}{25}=-\dfrac{15}{25}-\dfrac{20}{25}i=...$ $...=-\dfrac{3}{5}-\dfrac{4}{5}i$
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