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

1 Answer
Aug 6, 2016

Answer:

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#