#x_1+x_2=-1 " and " x_1 x_2=1#
#x_1+x_2=-1=>(x_1+x_2)^2=1=>x_1^2+x_2^2+2x_1 x_2=1#
#x_1^2 + x_2 ^2 + 2 \cdot 1=1=>x_1^2 + x_2^2 = -1#
So, by using sum and product we proof that the solutions are non real, because if #x_1# and #x_2"# were reals, the sum of their squares couldn't be negative.
#x_1 = a+b i#
#x_2 = a-b i#
(this came from the fact that if #a+bi# is a root, so #a-bi# is a root too, and the quadratic equation has two roots)
With a and b real, and b positive.
#x_1+x_2=-1=>a=-1/2#
So...
#x_1 = -1/2 + bi" and "x_1=-1/2-bi#
Lets use the product
#x_1 x_2 = 1=>(-1/2+bi)(-1/2-bi)=1#
#1/4+b^2=1=>b^2=3/4=>b=\sqrt(3)/2#
Then...
#x_1=(-1+i\sqrt(3))/2# and #x_2=(-1-i\sqrt(3))/2#
