# Given x1,x2,x3,x4 the solutions of the equation x^4+1=0 calculate the sum of the solutions and the sum of the squares of sol.

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x^4+1 = 0.

x1,x2,x3 and x4 are the roots of the equations.

Therefore x^4 +0x^3+0x^2 +0x+1 = 0.

Therefore by the relation between the roots and coeefficients of the equation we have:

Sum of the roots x1+x2+x3+x4 = -coefficient x^3/coefficient x^4 = 0.x1(x2+x3+x4)+x2(x3+x4)+x3x4 = coefficient of x^2 term in the equation/coefficient of x^4 . = 0/1 = 0.

Therefore x1^2+x^2+x^3+x4^2 = (x1+x2+x3+x4)^2 - 2{x1(x2+x3+x4)+x2(x3+x4)+x3x4} = 0^2-2*0 = 0.

Therefore x1+x2+x3+x4 = 0.

x1^2+x2^2+x3^2+x4^4 = 0.

We'll apply Viete expressions, which are the connection between roots and coefficients of an equation:

x1 + x2 + x3 + x4=0

x^2 + x2^2 + x3^2 + x4^2 = (x1 + x2 + x3 + x4)^2 - 2(x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4)

x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4= c/a=0/1=0

x^2 + x2^2 + x3^2 + x4^2 = 0 - 2*0=0