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

1 Answer
May 3, 2018

Answer:

#+-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)#