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 - Concept and Vocabulary Check - Page 314: 9


The statement $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$ simplifies to $x=-1\pm i\frac{\sqrt{6}}{2}$.

Work Step by Step

Consider the expression $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$. Simplify the expression. \[\begin{align} & x=\frac{-4\pm \sqrt{16-40}}{4} \\ & =\frac{-4\pm \sqrt{-24}}{4} \end{align}\] The principal square root of a negative number is such that for any positive real number $b$, $\sqrt{-b}=i\sqrt{b}$ Write the expression in terms of an imaginary unit $i$. \[x=\frac{-4\pm i\sqrt{24}}{4}\] Make the factors of $24$. \[x=\frac{-4\pm i\sqrt{4\cdot 6}}{4}\] Use the property of product of radicals $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$. \[\begin{align} & x=\frac{-4\pm i\sqrt{4}\cdot \sqrt{6}}{4} \\ & =\frac{-4\pm 2i\sqrt{6}}{4} \\ & =\frac{-4}{4}\pm \frac{2i\sqrt{6}}{4} \\ & =-1\pm i\frac{\sqrt{6}}{2} \end{align}\] Therefore, the correct fill for the blank is \[-1\pm i\frac{\sqrt{6}}{2}\] .
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.