Question

If the equation $(1+m^2)x^2+2mcx+(c^2-a^2)=0$ has equal roots, prove that $c^2=a^2(1+m^2).$

Answer 1

$$\begin{gathered} \text{Given:} \\ (1+m^2)x^2+2mcx+(c^2-a^2)=0 \\ \text{Here,} \\ a=(1+m^2),b=2mc\mathrm{and}c=(c^2-a^2) \\ \text{It is given that the roots of the equation are equal; therefore, we have:} \\ D=0 \\ \Rightarrow (b^2-4ac)=0 \\ \Rightarrow {\left(2mc\right)}^2-4\times (1+m^2)\times (c^2-a^2)=0 \\ \Rightarrow 4m^2{c}^2-4(c^2-a^2+m^2{c}^2-m^2{a}^2)=0 \\ \Rightarrow 4m^2{c}^2-4c^2+4a^2-4m^2{c}^2+4m^2{a}^2=0 \\ \Rightarrow -4c^2+4a^2+4m^2{a}^2=0 \\ \Rightarrow a^2+m^2{a}^2=c^2 \\ \Rightarrow a^2(1+m^2)=c^2 \\ \Rightarrow c^2=a^2(1+m^2) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$