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

Aug 31, 2015

No real zeros

#### Explanation:

The discriminant of $8 {x}^{2} - 4 x + 2 = 0$ is ${\left(- 4\right)}^{2} - 4 \cdot 8 \cdot 2$
= $16 - 64$ = -48

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

The function does have two complex zeros as follows:

x1 = $\frac{- \left(- 4\right) + \sqrt{- 48}}{2 \cdot 8}$
= $\frac{4 + \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot - 3}}{16}$
=$\frac{4 + 4 \cdot \sqrt{- 3}}{16}$
=$\frac{1 + \sqrt{3} \cdot i}{4}$ Where i = $\sqrt{- 1}$
Similarly, x2 = $\frac{1 - \sqrt{3} \cdot i}{4}$