Question

If the roots of the equation $\left(a^2+b^2\right)x^2-2\left(ac+bd\right)x+\left(c^2+d^2\right)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$.

Answer 1

It is given that the roots of the equation $\left(a^2+b^2\right)x^2-2\left(ac+bd\right)x+\left(c^2+d^2\right)=0$ are equal.
$$\begin{gathered} \therefore D=0 \\ \Rightarrow {\left[-2\left(ac+bd\right)\right]}^2-4\left(a^2+b^2\right)\left(c^2+d^2\right)=0 \\ \Rightarrow 4\left(a^2{c}^2+b^2{d}^2+2abcd\right)-4\left(a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\right)=0 \\ \Rightarrow 4\left(a^2{c}^2+b^2{d}^2+2abcd-a^2{c}^2-a^2{d}^2-b^2{c}^2-b^2{d}^2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(-a^2{d}^2+2abcd-b^2{c}^2\right)=0 \\ \Rightarrow -\left(a^2{d}^2-2abcd+b^2{c}^2\right)=0 \\ \Rightarrow {\left(ad-bc\right)}^2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow ad-bc=0 \\ \Rightarrow ad=bc \\ \Rightarrow \frac{a}{b}=\frac{c}{d} \end{gathered}$$
Hence Proved.