# Knowing that f(x) = x^4 + 1 determine S = x1 + x2 + x3 + x4 and T = x1^2 + x2^2 +x3^2 + x4^2

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f(x) = x^4+1

according Viete rule:

We know that x1+x2+x3+x4=-b/a= 0

Then S=0

(x1+x2+x3+x4)^2 = (x1+x2+x3+x4)(x1+x2+x3+x4)

Open brackets:

S^2 =x1^2+x1*x2+x1*x3+x1*x4+x2*x1+x2^2+x2*x3+x2*x4

+x3*x1+x3*x2+x3^2+x3*x4+x4*x1+x4*x2+x4*x3+x4^2

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

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

= T + 2(c/a)

==> T= s^2 - 2(c/a) = 0-0=0

x^4+1 = 0. To find x1+x2+x3+x4 and x1^2+x2^2+x3^2+x4^2 .

Solution:

x^4+1 = 0

By the relation between the roots and coefficients,

Sum of the roots = -coefficient of x^3/coefficient of x^4 = -0/1 = 0.

To find the sum x1^2+x2^2+x3^2+x4^2:

(x1+x2+x3+x4)^4 = (x1^2+x^2+x3^2+x4^2) +2[x1x2+x1x3+x1x4+x2x3+x2x4+x3x4).

So x1^2+x2^2+x3^2+x4^2 = (x1+x2+x3+x4)^2- 2(x1x2+x1x3+x1x4+x2x3+x3x4+x3x4) = 0 - 2*0, as the sum of the product of 2 different roots taken two at a time = (+1)*coefficient of x^2/coefficient of x^4 = 0/1 = 0

So, x1^2+x2^2+x3^2+x4^2 = 0

We'll use the Viete relations:

**S1 = 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=0/1=0

**S2 = x1^2 + x2^2 + x3^2 + x4^2 = 0 - 2*0 = 0**