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

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Answer 1 · 2015-08-31T02:44:16

No real zeros

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:

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 = $(1 - sqrt(3) * i)/4$

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