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: 48


The solution of the equation $2{{x}^{2}}+2x+3=0$ in standard form is $\left\{ -\frac{1}{2}+\frac{\sqrt{5}}{2}i,-\frac{1}{2}-\frac{\sqrt{5}}{2}i \right\}$.

Work Step by Step

Compare the equation $2{{x}^{2}}+2x+3=0$ with $a{{x}^{2}}+bx+c$. $\begin{align} & a=2 \\ & b=2 \\ & c=3 \end{align}$ Substitute $a=2$, $b=2$ and $c=3$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. $\begin{align} & x=\frac{-2\pm \sqrt{{{2}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\left( 2 \right)} \\ & =\frac{-2\pm \sqrt{4-24}}{4} \\ & =\frac{-2\pm \sqrt{-20}}{4} \end{align}$ Use the property $\sqrt{-b}=i\sqrt{b}$. \[\begin{align} & x=\frac{-2\pm i\sqrt{20}}{4} \\ & =\frac{-2\pm i\sqrt{4\cdot 5}}{4} \\ & =\frac{-2\pm 2i\sqrt{5}}{4} \end{align}\] Express the complex number in the standard form. \[\begin{align} & x=-\frac{2}{4}\pm \frac{2\sqrt{5}}{4}i \\ & =-\frac{1}{2}\pm \frac{\sqrt{5}}{2}i \end{align}\] Therefore, the solution of the equation $2{{x}^{2}}+2x+3=0$ in standard form is $\left\{ -\frac{1}{2}+\frac{\sqrt{5}}{2}i,-\frac{1}{2}-\frac{\sqrt{5}}{2}i \right\}$.
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