Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 366: 83


The solution for the equation ${{x}^{2}}+4x+6=0$ is $x=-2\pm \sqrt{2}i$

Work Step by Step

On comparing ${{x}^{2}}+4x+6=0$ with the standard form of the equation, $\begin{align} & a=1, \\ & b=4, \\ & c=6 \\ \end{align}$ Put these values in the formula: $\begin{align} & p=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 1\cdot 6}}{2\cdot 1} \\ & =\frac{-4\pm \sqrt{16-24}}{2} \\ & =\frac{-4\pm \sqrt{-8}}{2} \\ & =\frac{-4\pm \sqrt{8}\cdot \sqrt{-1}}{2} \end{align}$ And use the definition of the complex number i to obtain, $\begin{align} & p=\frac{-4\pm 2\sqrt{2}i}{2} \\ & =\frac{-4}{2}\pm \frac{2\sqrt{2}i}{2} \\ & =-2\pm \sqrt{2}i \end{align}$ Thus, the two solutions of the provided equations are $x=-2+\sqrt{2}i$ and $x=-2-\sqrt{2}i$.
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