How do you find the zeros, real and imaginary, of y=-4x^2+4x-16 using the quadratic formula?

Aug 6, 2016

2 imaginary roots:
$x = \frac{1 \pm i \sqrt{15}}{2}$

Explanation:

Use the improved quadratic formula in graphic form (Socratic Search)
$y = - 4 {x}^{2} + 4 x - 16 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 16 - 256 = - 240 < 0.$
Since D < 0, there are no real roots. There are 2 imaginary roots.
$D = 240 {i}^{2} = \left(16\right) \left(15\right) {i}^{2}$ --> $d = \pm 4 i \sqrt{15}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{4}{-} 8 \pm 4 i \frac{\sqrt{15}}{8} = \frac{1 \pm i \sqrt{15}}{2}$