# How do you find all the zeros of f(x)=x^4+3x^2-2?

Aug 5, 2016

$f \left(x\right)$ has zeros:

$\pm \sqrt{\frac{\sqrt{17} - 3}{2}}$

$\pm \sqrt{\frac{\sqrt{17} + 3}{2}} i$

#### Explanation:

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

$= {\left({x}^{2} + \frac{3}{2}\right)}^{2} - \frac{9}{4} - 2$

$= {\left({x}^{2} + \frac{3}{2}\right)}^{2} - \frac{17}{4}$

$= {\left({x}^{2} + \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{17}}{2}\right)}^{2}$

$= \left({x}^{2} + \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left({x}^{2} + \frac{3}{2} + \frac{\sqrt{17}}{2}\right)$

Hence zeros:

$x = \pm \sqrt{\frac{\sqrt{17} - 3}{2}}$

$x = \pm \sqrt{\frac{\sqrt{17} + 3}{2}} i$