Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is

$\frac{m}{n} x^{2}+\frac{n}{m}=1-2 x$

$\Rightarrow \frac{m^{2} x^{2}+n^{2}}{m n}=1-2 x$

$\Rightarrow m^{2} x^{2}+n^{2}=m n-2 m n x$

$\Rightarrow m^{2} x^{2}+2 m n x+n^{2}-m n=0$

This equation is of the form $a x^{2}+b x+c=0$, where $a=m^{2}, b=2 m n$ and $c=n^{2}-m n$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(2 m n)^{2}-4 \times m^{2} \times\left(n^{2}-m n\right)=4 m^{2} n^{2}-4 m^{2} n^{2}+4 m^{3} n=4 m^{3} n>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4 m^{3} n}=2 m \sqrt{m n}$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-2 m n+2 m \sqrt{m n}}{2 \times m^{2}}=\frac{2 m(-n+\sqrt{m n})}{2 m^{2}}=\frac{-n+\sqrt{m n}}{m}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-2 m n-2 m \sqrt{m n}}{2 \times m^{2}}=\frac{-2 m(n+\sqrt{m n})}{2 m^{2}}=\frac{-n-\sqrt{m n}}{m}$

Hence, $\frac{-n+\sqrt{m n}}{m}$ and $\frac{-n-\sqrt{m n}}{m}$ are the roots of the given equation.