## Precalculus (6th Edition) Blitzer

The solution set for the quadratic equation, ${{x}^{2}}-2x+4=0$ is $\left\{ 1\pm i\sqrt{3} \right\}$.
Consider the quadratic equation, ${{x}^{2}}-2x+4=0$ Compare the provided equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$. Here, $a=1,\text{ }b=-2\text{ and }c=4$ Apply the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Substitute 1 for a, $-2$ for b and 4 for c. $x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}$ Simplify the radical. \begin{align} & x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)} \\ & =\frac{2\pm \sqrt{4-16}}{2} \\ & =\frac{2\pm \sqrt{-12}}{2} \end{align} As $i=\sqrt{-1}$ \begin{align} & x=\frac{2\pm \sqrt{12}\sqrt{-1}}{2} \\ & =\frac{2\pm 2i\sqrt{3}}{2} \\ & =\frac{2\left( 1\pm i\sqrt{3} \right)}{2} \\ & =1\pm i\sqrt{3} \end{align} The solutions of the quadratic are complex conjugates of each other. Hence, the solution set for the quadratic equation ${{x}^{2}}-2x+4=0$ is $\left\{ 1\pm i\sqrt{3} \right\}$.