# How do you find the roots, real and imaginary, of y=-3(x -1 )^2-3x-1 using the quadratic formula?

Apr 7, 2018

The roots are $x = \frac{1}{2} + i \frac{\sqrt{39}}{6}$ and $x = \frac{1}{2} - i \frac{\sqrt{39}}{6}$.

#### Explanation:

First, express $y$ in standard form.

$y = \textcolor{red}{- 3 {\left(x - 1\right)}^{2}} - 3 x - 1 = \textcolor{red}{- 3 {x}^{2} + 6 x - 3} - 3 x - 1 = - 3 {x}^{2} + 3 x - 4$

The quadratic formula finds when $y = 0$. $y = 0$ when

$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(- 4\right) \left(- 3\right)}}{2 \cdot \left(- 3\right)}$

$= \frac{- 3 \pm \sqrt{9 - 48}}{2 \cdot \left(- 3\right)} = \frac{- 3 \pm \sqrt{- 39}}{2 \cdot \left(- 3\right)} = \frac{- 3 \pm i \sqrt{39}}{2 \cdot \left(- 3\right)}$

$= \frac{1}{2} \pm i \frac{\sqrt{39}}{6}$

Therefore the roots are

$x = \frac{1}{2} + i \frac{\sqrt{39}}{6}$ and $x = \frac{1}{2} - i \frac{\sqrt{39}}{6}$.