# What kind of equation is z^4-3z^2+2=0 ?

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The equation you have given is z^4 - 3z^2 + 2 = 0. Here you can replace z^2 = x to get a quadratic equation in x which can be solved.

z^4 - 3z^2 + 2 = 0

=> x^2 - 3x + 2 = 0

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

=> x(x - 2) - 1(x - 2) = 0

=> (x - 1)(x - 2) = 0

So we have x = 1 and x = 2

As x = z^2

z^2 = 1 and z^2 = 2

z = -1 and z = +1 and z = +sqrt 2 and - sqrt 2

**The roots of the equation are -1 , 1 , sqrt 2 and -sqrt 2**

This equation is called biquadratic equation and it is easy to solve.

This equation is reduced to a quadratic equation when doing the substitution z^2 = x.

We'll re-write the equation in x:

x^2 - 3x + 2 = 0

We'll apply quadratic formula:

x1 = [3+ sqrt(9-8)]/2

x1 = (3+1)/2

x1 = 2

x2 = (3-1)/2

x2 = 1

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

z^2 = x1

z^2 = 2

z1 = sqrt 2 and z2 = -sqrt 2

z^2 = x2

z^2 = 1

z3 = -1 and z4 = 1

**The solutions of the biquadraticÂ equation are: {-sqrt2 ; -1 ; 1 ; sqrt2}.**