Question

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

Answer 1

`+-sqrt(5/2)`
`+-sqrt(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+bi` & `a-bi`
2. a pair of the complex conjugate of `a+bi` & `a-bi` and two real roots
3. four real roots

`4x^4-16x^2+15=0`
first I guess I can use "Cross method" to factorizative this equation
it can be seen as below
`(2x^2-5)(2x^2-3)=0`
so there are four real roots
`+-sqrt(5/2)`
`+-sqrt(3/2)`