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

1 Answer
May 3, 2018

+-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)