Answer:
[tex]S = -\frac{b}{a}[/tex]
Step-by-step explanation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex]
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this problem, the sum S of the solutions of a quadratic equation is:
[tex]S = x_{1} + x_{2}[/tex]
So
[tex]S = \frac{-b + \sqrt{\bigtriangleup}}{2*a} + \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
They have the same denominators, so we can keep the denominators and sum the numerators.
[tex]S = \frac{-b + \sqrt{\bigtriangleup} - b - \sqrt{\bigtriangleup}}{2a}[/tex]
[tex]S = \frac{-2b}{2a}[/tex]
[tex]S = -\frac{b}{a}[/tex]
