Question

B. x4 + 10x2 + 25 = 0

Answer 1

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