## Precalculus (6th Edition) Blitzer

The statement $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$ simplifies to $x=-1\pm i\frac{\sqrt{6}}{2}$.
Consider the expression $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$. Simplify the expression. \begin{align} & x=\frac{-4\pm \sqrt{16-40}}{4} \\ & =\frac{-4\pm \sqrt{-24}}{4} \end{align} The principal square root of a negative number is such that for any positive real number $b$, $\sqrt{-b}=i\sqrt{b}$ Write the expression in terms of an imaginary unit $i$. $x=\frac{-4\pm i\sqrt{24}}{4}$ Make the factors of $24$. $x=\frac{-4\pm i\sqrt{4\cdot 6}}{4}$ Use the property of product of radicals $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$. \begin{align} & x=\frac{-4\pm i\sqrt{4}\cdot \sqrt{6}}{4} \\ & =\frac{-4\pm 2i\sqrt{6}}{4} \\ & =\frac{-4}{4}\pm \frac{2i\sqrt{6}}{4} \\ & =-1\pm i\frac{\sqrt{6}}{2} \end{align} Therefore, the correct fill for the blank is $-1\pm i\frac{\sqrt{6}}{2}$ .