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 314: 46

Answer

The solution of the equation ${{x}^{2}}-2x+17=0$ in standard form is $\left\{ 1+4i,1-4i \right\}$.

Work Step by Step

Consider the equation,${{x}^{2}}-2x+17=0$ Compare the equation ${{x}^{2}}-2x+17=0$ with $a{{x}^{2}}+bx+c$. $\begin{align} & a=1 \\ & b=-2 \\ & c=17 \end{align}$ Substitute $a=1$, $b=-2$ and $c=17$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. $\begin{align} & x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 17 \right)}}{2\left( 1 \right)} \\ & =\frac{2\pm \sqrt{4-68}}{2} \\ & =\frac{2\pm \sqrt{-64}}{2} \end{align}$ Use the property $\sqrt{-b}=i\sqrt{b}$. \[\begin{align} & x=\frac{2\pm i\sqrt{64}}{2} \\ & =\frac{2\pm 8i}{2} \\ & =\frac{2}{2}\pm \frac{8}{2}i \\ & =1\pm 4i \end{align}\] Therefore, the solution of the equation ${{x}^{2}}-2x+17=0$ in standard form is $\left\{ 1+4i,1-4i \right\}$.
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.