## Precalculus (6th Edition) Blitzer

The solution of the expression is $\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$.
We have to convert the equation into the general form; subtract ${{x}^{2}}+x$ from both sides to get: \begin{align} & 3{{x}^{2}}+1-{{x}^{2}}-x={{x}^{2}}+x-{{x}^{2}}-x \\ & 2{{x}^{2}}-x+1=0 \end{align} Here, $a=2$ , $b=-1$ and $c=1$. By using the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get the solution, \begin{align} & x=\frac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 2 \right)\left( 1 \right)}}{2\left( 2 \right)} \\ & =\frac{1\pm \sqrt{1-8}}{4} \\ & =\frac{1\pm \sqrt{-7}}{4} \end{align} Since, ${{i}^{2}}=-1$ , Then, \begin{align} & x=\frac{1\pm \sqrt{7\left( {{i}^{2}} \right)}}{4} \\ & =\frac{1\pm i\sqrt{7}}{4} \\ & =\frac{1}{4}\pm i\frac{\sqrt{7}}{4} \end{align} Now, to check whether the solution is correct or not, put $x=\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$ in the equation and get: \begin{align} & 3{{\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right)}^{2}}+1={{\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right)}^{2}}+\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right) \\ & 3\left( \frac{1\pm 2i\sqrt{7}-7}{16} \right)+\frac{16}{16}=\frac{1\pm 2i\sqrt{7}-7}{16}+\frac{4\left( 1\pm i\sqrt{7} \right)}{16} \\ & \frac{3\pm 6i\sqrt{7}-21+16}{16}=\frac{\pm 2i\sqrt{7}-6+4\pm i4\sqrt{7}}{16} \\ & \frac{-2\pm 6i\sqrt{7}}{16}=\frac{-2\pm 6i\sqrt{7}}{16} \end{align} Thus, the solution of the given expression is $\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$.
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