This equation is also called biquadratic equation.

To solve this equation, we'll reduce it to a quadratic equation by replacing z^2 = x.

We'll re-write the equation in x:

x^2 - 12x + 11 = 0

We'll apply quadratic formula:

x1 = [12+ sqrt(144-44)]/2

x1 = (12+10)/2

x1 = 11

x2 = (12-10)/2

x2 = 1

But, we'll have to find z1,z2,z3,z4.

z^2 = x1

z^2 = 11

z1 = sqrt 11 and z2 = -sqrt 11

z^2 = x2

z^2 = 1

z3 = -1 and z4 = 1

**The solutions of the biquadraticÂ equation are: {-sqrt11 ; -1 ; 1 ; sqrt11}.**

The equation z^4-12z^2+11=0 has to be solved for z.

Let x = z^2, the given equation is now:

x^2 - 12x + 11 = 0

x^2 - 11x - x + 11 = 0

x(x - 11) - 1(x - 11) = 0

(x - 1)(x - 11) = 0

x = 1 and x = 11

As x = z^2

z^2 = 1

z = `+-1`

z^2 = 11

z = `+- sqrt 11`

The solution of the equation is `+-1` and `+-sqrt 11`