# How do you find the zeros of y=4x^4-11x^2-3?

Sep 22, 2016

$x = \pm \frac{1}{2} i$

$x = \pm \sqrt{3}$

#### Explanation:

$4 {x}^{4} - 11 {x}^{2} - 3 = 0$ can be factored

$\left(4 {x}^{2} + 1\right) \left({x}^{2} - 3\right) = 0$

So we have

$4 {x}^{2} + 1 = 0$ or ${x}^{2} - 3 = 0$

Proceeding solve each factor for $x$

$4 {x}^{2} = - 1$

${x}^{2} = - \frac{1}{4}$

$x = \pm \sqrt{\frac{1}{4}} i$

$x = \pm \frac{1}{2} i$

${x}^{2} = 3$

$x = \pm \sqrt{3}$