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

Dec 26, 2016

There are a double rooth, x = - 1.

#### Explanation:

Applying the Quadratic Formula, we can obtain the rooth of:

$y = - {x}^{2} - 2 x - 1$,

solving the equation:

$- {x}^{2} - 2 x - 1 = 0$.

When $a {x}^{2} + b x + c = 0$, the solutions are:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$.

Applying this formula to the equation given, we have:

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

$= \frac{2 \pm \sqrt{4 - 4}}{- 2} = - 1$.

Because the discriminant is zero, we have only one solution. In this case, the solution is real.