# How do you find the zeros, real and imaginary, of y=-x^2-x-9 using the quadratic formula?

Mar 25, 2018

#### Answer:

$x = - \frac{1}{2} \pm \frac{1}{2} \sqrt{35} \textcolor{w h i t e}{.} i$

#### Explanation:

You have two ways of tackling this. You may complete the square or use 'the formula'

I choose to use the formula. It really is worth trying to commit this one to memory. It crops up quite often.

Standard form $y = a {x}^{2} + b x + c \textcolor{w h i t e}{\text{dd") -> color(white)("dd}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Given $y = - {x}^{2} - x - 9$

where a=-1;b=-1 and c=-9

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(- 1\right) \left(- 9\right)}}{2 \left(- 1\right)}$

$x = \frac{+ 1 \pm \sqrt{- 35}}{- 2}$

$x = - \frac{1}{2} \pm \frac{\sqrt{35} \times \sqrt{- 1}}{2}$

$x = - \frac{1}{2} \pm \frac{1}{2} \sqrt{35} \textcolor{w h i t e}{.} i$