If the equation \[\left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0\] has equal roots, then prove that \[{{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right)\].

Hint: Given equation is a quadratic equation, compare it with general quadratic equation and get values of a, b and c. Given discriminant is zero, as it has equal roots. Hence substitute a, b and c in the discriminant of the quadratic formula which is given by \[D={{b}^{2}}-4ac\] for quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] and prove it.

Complete step-by-step answer:
We have been given a quadratic equation, which we need to solve and prove that, \[{{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right)\].
The quadratic equation given to us is,
\[\Rightarrow \left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0\] - (1)
We know that the general quadratic equation is represented as,
\[\Rightarrow a{{x}^{2}}+bx+c=0\] - (2)
By comparing both equations (1) and (2), we get,
\[\Rightarrow a=\left( 1+{{m}^{2}} \right)\], \[b=2mc\], \[c={{c}^{2}}-{{a}^{2}}\]
We know the quadratic formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
In this formula \[\left( {{b}^{2}}-4ac \right)\] is called discriminant. This determines the number of roots of solutions in a quadratic equation. We can make cases like,
If \[{{b}^{2}}-4ac<0\], the equation has zero real solutions.
If \[{{b}^{2}}-4ac=0\], the equation has 1 real solution.
If \[{{b}^{2}}-4ac>0\], the equation has 2 real solutions.
Discriminant is represented as D.
\[\therefore D={{b}^{2}}-4ac\] - (3)
Now from equation (1), the given quadratic equation is equal to zero. Thus we can say that equation (1) has equal roots. Thus we can put, D = 0 in equation (3).
Let us put D = 0.
\[\begin{align}
  & \therefore D={{b}^{2}}-4ac \\
 & 0={{b}^{2}}-4ac \\
 & \Rightarrow {{b}^{2}}=4ac \\
\end{align}\]
Now we can, \[a=1+{{m}^{2}}\],\[b=2mc\] and \[c={{c}^{2}}-{{a}^{2}}\]. Thus the above expression becomes,
\[\Rightarrow {{b}^{2}}=4ac\]
\[\Rightarrow {{\left( 2mc \right)}^{2}}=4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)\], Expand and simplify it,
\[\begin{align}
  & \Rightarrow 4{{m}^{2}}{{c}^{2}}=4\left[ {{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{a}^{2}}{{m}^{2}} \right] \\
 & \Rightarrow 4{{m}^{2}}{{c}^{2}}=4{{c}^{2}}-4{{a}^{2}}+4{{m}^{2}}{{c}^{2}}-4{{a}^{2}}{{m}^{2}} \\
\end{align}\]
Cancelling out \[4{{m}^{2}}{{c}^{2}}\] from LHS and RHS we get,
\[\begin{align}
  & \therefore 4{{c}^{2}}-4{{a}^{2}}-4{{a}^{2}}{{m}^{2}}=0 \\
 & \Rightarrow 4{{c}^{2}}=4\left( {{a}^{2}}+{{a}^{2}}{{m}^{2}} \right) \\
\end{align}\]
Cancelling out 4 from LHS and RHS we get,
\[\Rightarrow {{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right)\]
Thus we proved that the equation, \[\left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0\] has equal roots and we get, \[{{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right)\].

Note: Some students might directly substitute the values of a, b, c in the quadratic formula to find the roots. This is a very common thing that comes to the mind of students. But they must note here that doing this will not give the required proof. So, the condition for equal roots, i.e. D = 0 must be remembered and made use of to solve this question.