Solve the equation x^4+4=0

x^4+4=0
=>x^4+4x^2+4=4x^2

=> (x^2+2)^2=4x^2

=> x^2+2= 2x, -2x

=>x^2+2x+2=0 & x^2-2x+2=0

by using the formule of solving a quadratic equation (ax^2+bx+c=0)
is 1/2a times -b+square root of(b^-4ac)
1/2a times -b-square root of(b^2-4ac)


totally we get 4 'x' values ie..,-1+i , -1-i ,1+i , 1-i
where i=square root of -1

Solve the equation x^4+4=0?
Sol x^4 = -4 = (2i)^2
taking square root
x^2 = 2i or x^2 = -2i.
x = sqrt(2i) or x = - sqrt(2i) and x^2 = -2i...............Ans
But there is no real roots

There is no solution.
x^4 will always be a positive nimber.
A positive number plus 4 added to it can never be negative.

If Complex number solutions are allowed, then that's perhaps a different matter