How many zeroes does the function have #y=8x^2-4x+2#?

1 Answer
Aug 31, 2015

No real zeros

Explanation:

The discriminant of #8x^2 - 4x + 2 = 0# is #(-4)^2 - 4*8 *2#
= #16 - 64# = -48

Since the discriminant < 0 the function has no real zeros.

The function does have two complex zeros as follows:

Using the Quadratic Formula

x1 = #(-(-4) + sqrt( -48))/(2*8)#

= #(4+sqrt(2*2*2*2*-3))/16#

=#(4+4*sqrt(-3))/16#

=#(1 + sqrt(3) * i)/4# Where i = #sqrt(-1)#

Similarly, x2 = #(1 - sqrt(3) * i)/4#