Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 315: 60


The simplified form of the expression $\frac{{{x}^{2}}+11}{3-x}$ for $x=4i$ is $-\frac{3}{5}-\frac{4}{5}i$.

Work Step by Step

Consider the expression, $\frac{{{x}^{2}}+11}{3-x}$ Substitute $x=4i$ in the expression $\frac{{{x}^{2}}+11}{3-x}$. $\begin{align} & \frac{{{x}^{2}}+11}{3-x}=\frac{{{\left( 4i \right)}^{2}}+11}{3-\left( 4i \right)} \\ & =\frac{16{{i}^{2}}+11}{3-4i} \end{align}$ Replace the value ${{i}^{2}}=-1$. $\begin{align} & \frac{{{x}^{2}}+11}{3-x}=\frac{16\left( -1 \right)+11}{3-4i} \\ & =\frac{-16+11}{3-4i} \\ & =\frac{-5}{3-4i} \end{align}$ Multiply by complex conjugate of the denominator in the numerator and the denominator. \[\frac{{{x}^{2}}+11}{3-x}=\frac{-5}{\left( 3-4i \right)}\cdot \frac{\left( 3+4i \right)}{\left( 3+4i \right)}\] Distribute $-5$ within the parentheses in numerator and use the FOIL method in the denominator. \[\begin{align} & \frac{{{x}^{2}}+11}{3-x}=\frac{-5\left( 3+4i \right)}{\left( 3-4i \right)\left( 3+4i \right)} \\ & =\frac{-15-20i}{9+12i-12i-16{{i}^{2}}} \\ & =\frac{-15-20i}{9-16{{i}^{2}}} \end{align}\] Replace the value ${{i}^{2}}=-1$. \[\begin{align} & \frac{{{x}^{2}}+11}{3-x}=\frac{-15-20i}{9-16\left( -1 \right)} \\ & =\frac{-15-20i}{25} \\ & =-\frac{15}{25}-\frac{20}{25}i \\ & =-\frac{3}{5}-\frac{4}{5}i \end{align}\] Therefore, the simplified form of the expression $\frac{{{x}^{2}}+11}{3-x}$ for $x=4i$ is $-\frac{3}{5}-\frac{4}{5}i$.
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