B. x4 + 10x2 + 25 = 0

Solve it like a normal quadratic equation and take the root of the roots

x^4 + 10x^2 + 25 = 0

Δ = b² - 4. a. c

Δ = 10² - 4. 1. 25

Δ = 100 - 4. 1. 25

Δ = 0

1 real root.

In this case, x' = x'':

x = (-b + - √Δ) / 2a

x' = (-10 + √0) / 2.1

x'' = (-10 - √0) / 2.1

x' = - 10 / 2

x'' = - 10 / 2

x' = - 5

x'' = - 5

Now take the root of this roots

√-5 = i√5

The root of this equation is x = ±i√5