Answer
Proved below,
Work Step by Step
For the equation $a{{x}^{2}}+bx+c=0$, use the quadratic equation to solve.
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Thus the two solutions are,
$x=\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$
Or,
$x=\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
The sum of the solutions is,
$\begin{align}
& \left( \frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right)+\left( \frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right)=\frac{-b+\sqrt{{{b}^{2}}-4ac}-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-2b}{2a} \\
& =\frac{-b}{a}
\end{align}$
Thus, the sum of the solutions of the equation $a{{x}^{2}}+bx+c=0$ is $-\frac{b}{a}$.
