Precalculus (6th Edition) Blitzer

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

Chapter 2 - Mid-Chapter Check Point - Page 381: 7


The solution set for the quadratic equation, $x\left( 2x-3 \right)=-4$ is $\left\{ \frac{3}{4}\pm \frac{\sqrt{23}}{4}i \right\}$.

Work Step by Step

Consider quadratic equation, $x\left( 2x-3 \right)=-4$ Convert it to the standard form. Add 4 to both sides and multiply x with terms in bracket. $\begin{align} & x\left( 2x-3 \right)+4=-4+4 \\ & 2{{x}^{2}}-3x+4=0 \end{align}$ Compare the equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$. Here, $a=2,\text{ }b=-3\text{ and }c=4$ Apply quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Substitute 2 for a, $-3$ for b and 4 for c. $x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\left( 2 \right)}$ Simplify the radical. $\begin{align} & x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\left( 2 \right)} \\ & =\frac{3\pm \sqrt{9-32}}{4} \\ & =\frac{3\pm \sqrt{-23}}{4} \end{align}$ Rewrite $\frac{3\pm \sqrt{-23}}{4}$ as $\frac{3\pm \sqrt{23}\sqrt{-1}}{4}$ As $i=\sqrt{-1}$ Therefore, $\begin{align} & x=\frac{3\pm i\sqrt{23}}{4} \\ & =\frac{3}{4}\pm \frac{\sqrt{23}}{4}i \end{align}$ The solutions of the quadratic are complex conjugates of each other. Hence, the solution set for the quadratic equation $x\left( 2x-3 \right)=-4$ is $\left\{ \frac{3}{4}\pm \frac{\sqrt{23}}{4}i \right\}$.
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