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