Question

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

Answer 1

2 imaginary roots:
`x = (1 +- isqrt15)/2`

Explanation:

Use the improved quadratic formula in graphic form (Socratic Search)
`y = -4x^2 + 4x - 16 = 0`
`D = d^2 = b^2 - 4ac = 16 - 256 = -240 < 0.`
Since D < 0, there are no real roots. There are 2 imaginary roots.
`D = 240i^2= (16)(15)i^2` --> `d = +- 4isqrt15`
`x = -b/(2a) +- d/(2a) = -4/-8 +- 4isqrt15/8 = (1 +- isqrt15)/2`