# How do you solve 4x^4 - 16x^2 + 15 = 0?

May 3, 2018

$\pm \sqrt{\frac{5}{2}}$
$\pm \sqrt{\frac{3}{2}}$

#### Explanation:

for real coefficient equation
equation of n‐th degree exist n roots
so this equations exists 3 possible answers
1. two pairs of the complex conjugate of $a + b i$ & $a - b i$
2. a pair of the complex conjugate of $a + b i$ & $a - b i$ and two real roots
3. four real roots

$4 {x}^{4} - 16 {x}^{2} + 15 = 0$
first I guess I can use "Cross method" to factorizative this equation
it can be seen as below
$\left(2 {x}^{2} - 5\right) \left(2 {x}^{2} - 3\right) = 0$
so there are four real roots
$\pm \sqrt{\frac{5}{2}}$
$\pm \sqrt{\frac{3}{2}}$