question_answer

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

Answer:

The given equation \[(1+{{m}^{2}}){{x}^{2}}+2mcx+{{c}^{2}}-{{a}^{2}}=0\] has equal roots

Here, \[A=1+{{m}^{2}},B=2mc,C={{c}^{2}}-{{a}^{2}}\]

For equal roots, \[~D=0={{B}^{2}}-4AC\]

\[\Rightarrow \]   \[{{(2mc)}^{2}}-4(1+{{m}^{2}})({{c}^{2}}-{{a}^{2}})=0\]

\[\Rightarrow \]   \[4{{m}^{2}}{{c}^{2}}-4({{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}})=0\]

\[\Rightarrow \]   \[{{m}^{2}}{{c}^{2}}-{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{2}}=0\]

\[\Rightarrow \]   \[-{{c}^{2}}+{{a}^{2}}(1+{{m}^{2}})=0\]

\[\Rightarrow \]               \[{{c}^{2}}={{a}^{2}}(1+{{m}^{2}})\]                  Hence Proved.