Question

show that the equation x^6-3x^2-x+1=0 has at least two imaginary roots​

Answer 1

Answer:

The signs in the equation changes two times then it has two imaginary roots

Answer 2

It has max 2 positive real roots and max 2 negative real roots hence at least 2 imaginary roots.

The number of time the sign changes in f(x) are the number of maximum positive roots.

Here f(x) = x^6-3x^2-x+1.

Here sign changes twice => max 2 positive roots.

The number of time the sign changes in f(-x) are the number of maximum negative roots.

Hence f(-x) = x^6-3x^2+x+1.

Here sign changes twice => max 2 negative roots.

Hence the min number of imaginary roots = total - max of rest = 6-4 =2.