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: 59

Answer

The simplified form of the expression $\frac{{{x}^{2}}+19}{2-x}$ for $x=3i$ is $\frac{20}{13}+\frac{30}{13}i$.

Work Step by Step

Consider the expression, $\frac{{{x}^{2}}+19}{2-x}$ Substitute $x=3i$ in the expression $\frac{{{x}^{2}}+19}{2-x}$. $\begin{align} & \frac{{{x}^{2}}+19}{2-x}=\frac{{{\left( 3i \right)}^{2}}+19}{2-\left( 3i \right)} \\ & =\frac{9{{i}^{2}}+19}{2-3i} \end{align}$ Replace the value ${{i}^{2}}=-1$. $\begin{align} & \frac{{{x}^{2}}+19}{2-x}=\frac{9\left( -1 \right)+19}{2-3i} \\ & =\frac{-9+19}{2-3i} \\ & =\frac{10}{2-3i} \end{align}$ Multiply by complex conjugate of the denominator in the numerator and the denominator. \[\frac{{{x}^{2}}+19}{2-x}=\frac{10}{\left( 2-3i \right)}\cdot \frac{\left( 2+3i \right)}{\left( 2+3i \right)}\] Distribute $10$ within the parentheses in numerator and use the FOIL method in the denominator. \[\begin{align} & \frac{{{x}^{2}}+19}{2-x}=\frac{10\left( 2+3i \right)}{\left( 2-3i \right)\left( 2+3i \right)} \\ & =\frac{20+30i}{4+6i-6i-9{{i}^{2}}} \\ & =\frac{20+30i}{4-9{{i}^{2}}} \end{align}\] Replace the value ${{i}^{2}}=-1$. \[\begin{align} & \frac{{{x}^{2}}+19x}{2-x}=\frac{20+30i}{4-9\left( -1 \right)} \\ & =\frac{20+30i}{4+9} \\ & =\frac{20+30i}{13} \\ & =\frac{20}{13}+\frac{30}{13}i \end{align}\] Therefore, the simplified form of the expression $\frac{{{x}^{2}}+19}{2-x}$ for $x=3i$ is $\frac{20}{13}+\frac{30}{13}i$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.