The solutions of the equation x^4 + 6x^2 + 5 = 0 have to be determined. It can be seen that the highest power of x in the equation is 4. As a result the equation has 4 solutions.

x^4 + 6x^2 + 5 = 0

Let x^2 = y

=> y^2 + 6y + 5 = 0

=> y^2 + 5y + y + 5 = 0

=> y(y + 5) + 1(y + 5) = 0

=> (y + 1)(y + 5) = 0

y + 1 = 0

=> y = -1

y + 5 = 0

=> y = -5

As y = x^2

x^2 = -1

=> x = i and x = -i

x^2 = -5

=> x = `i*sqrt 5` and x = `-sqrt 5*i`

**The solutions of the equation x^4 + 6x^2 + 5 = 0 are **`{-i, -sqrt 5*i, sqrt 5*i, i}`

Given :- x^4 + 6x^2 + 5 = 0

=> (x^2)^2 + 6(x^2) + 5 = 0

Let x^2 be = a , then the given equation

(x^2)^2 + 6(x^2) + 5 = a^2 + 6a + 5 = 0

Taking the equation :- a^2 + 6a + 5 =0

=> a^2 + 5a + a + 5 = 0

=> a(a+5)+1(a+5) = 0

=> (a+5)(a+1) = 0

case I :

If a+5 = 0 then, a = -5

Or, x^2= -5 [ substituting value of 'a' ]

Or, x = sqrt(-5)

x = √5 i and x = - √5 i [ x will be imaginary number]

case II :

If a + 1 =0 then , a = -1

Or, x^2 = -1

Or, x = sqrt(-1)

Or, x = i and x = -i [ x will be imaginary number] Hence ,

x = [ i,-i , √5 i, -√5 i ] <--- Answer