How do you find the zeros, real and imaginary, of y=-x^2-4x-11 using the quadratic formula?

1 Answer
Jul 4, 2017

2 imaginary roots:
$x = - 2 \pm i \sqrt{7}$

Explanation:

$y = - {x}^{2} - 4 x - 11 = 0$
Use the improved quadratic formula in graphic form (Google Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 16 - 44 = - 28$
Since D < 0, there are 2 imaginary roots.
$d = \pm 2 i \sqrt{7}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{-} 2 \pm 2 i \frac{\sqrt{7}}{2} = - 2 \pm i \sqrt{7}$