# Solve the equation t^4+10t-11=0

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It is not possible to solve the equation you have given t^4 + 10t - 11 = 0.

Instead, here is the solution for t^4 + 10 t^2 - 11 = 0

let t^2 = x

=> x^2 + 10x - 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

t^2 = 1

=> t1 = 1

=> t2 = -1

t^2 = -11

=> t3 = i*sqrt 11

=> t4 = -i*sqrt 11

**The solutions of the equation t^4 + 10 t^2 - 11 = 0 are {-1 , 1, i*sqrt 11, -i*sqrt 11}**

Thiskind of equation is called biquadratic equation.

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

We'll re-write the equation in x:

x^2 +10x - 11 = 0

We'll apply quadratic formula:

x1 = [-10+ sqrt(100+44)]/2

x1 = (-10+12)/2

x1 = 1

x2 = (-10-12)/2

x2 = -11

But, we'll have to find t1,t2,t3,t4.

t^2 = x1

t^2 = 1

t1 = sqrt 1 and t2 = -sqrt 1

t1=1 and t2=-1

t^2 = x2

t^2 = -11

z3 = i*sqrt11 and z4 = -i*sqrt11

**The solutions of the biquadratic equation are real and complex: { -1 ; 1 } ; {-i*sqrt11 ; i*sqrt11}.**