Question

If the equation (1+m square) x square +2m cx+(c square-a square)=0 has equal roots, than prove that c square=a square (1+m square)

Answer 1

here is your answer guys
ot should be applied by discriminant formula

Answer 2

$\bold{c^{2}=a^{2}\left(1+m^{2}\right)}$ if the equation $\bold{\left(1+m^{2}\right) \times x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0}$ has equal roots.

Given:

The equation $\left(1+m^{2}\right) \times x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$has equal roots.

To prove:

Prove that $c^{2}=a^{2}\left(1+m^{2}\right)$

Proof:

Any quadratic equation $a x^{2}+b x+c=0$ has real roots when the discriminant is equal to 0. That is  

$b^{2}-4 a c=0$

Consider the given equation $\left(1+m^{2}\right) \times x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$

This equation is a quadratic equation in x.

Here  $b=2 m c ; a=1+m^{2} ; c=c^{2}-a^{2}$

So, $b^{2}-4 a c=(2 m c)^{2}-4\left(1+m^{2}\right)\left(c^{2}-a^{2}\right)=0$

$\begin{array}{l}{\Rightarrow 4 m^{2} \times c^{2}-4\left(c^{2}+m^{2} \times c^{2}-a^{2}-m^{2} \times a^{2}\right)=0} \\ {\Rightarrow m^{2} \times c^{2}-c^{2}-m^{2} \times c^{2}+a^{2}+m^{2} \times a^{2}=0} \\ {\Rightarrow c^{2}-a^{2}=m^{2} \times a^{2}}\end{array}$

$\Rightarrow c^{2}=a^{2}\left(1+m^{2}\right)$

Hence proved.  

$\bold{c^{2}=a^{2}\left(1+m^{2}\right)}$ if the equation $\bold{\left(1+m^{2}\right) \times x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0}$ has equal roots.