# How do you find the number of complex, real and rational roots of 2x^4-18x^2+5=0?

Nov 8, 2016

Find the zeros.

#### Explanation:

First substitute $a$ for ${x}^{2}$.
$2 {a}^{2} - 18 a + 5 = 0$

Next find what $a$ is using the quadratic formula:
$a = \setminus \frac{18 \setminus \pm \setminus \sqrt{{\left(- 18\right)}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$

Next, simplify:
$a = \setminus \frac{18 \setminus \pm \setminus \sqrt{324 - 40}}{4}$
$a = \setminus \frac{18 \setminus \pm \setminus \sqrt{284}}{4}$
$a = \setminus \frac{18 \setminus \pm 2 \setminus \sqrt{71}}{4}$
$a = \setminus \frac{9 \setminus \pm \setminus \sqrt{71}}{2}$

$a = {x}^{2}$, so:
${x}^{2} = \setminus \frac{9 \setminus \pm \setminus \sqrt{71}}{2}$
$x = \setminus \pm \setminus \sqrt{\setminus \frac{9 \setminus \pm \setminus \sqrt{71}}{2}}$

So the four zeros are:
$\setminus \sqrt{\setminus \frac{9 - \setminus \sqrt{71}}{2}} , \setminus \sqrt{\setminus \frac{9 + \setminus \sqrt{71}}{2}} , - \setminus \sqrt{\setminus \frac{9 - \setminus \sqrt{71}}{2}} , - \setminus \sqrt{\setminus \frac{9 + \setminus \sqrt{71}}{2}}$

$\sqrt[]{71}$ is less than 9, so all of the zeros are real. $\sqrt{71}$ is irrational, so none of the zeros will be rational.