The roots of the quadratic equation $\frac{m}{n}{x}^2+\frac{n}{m}=1-2x$ is

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

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

$\Rightarrow m^2{x}^2+2mnx+(n^2-mn)=0$

$\Rightarrow (m^2{x}^2+mnx+m\sqrt{mnx})+(mnx-m\sqrt{mnx}(n+\sqrt{mn})(n-\sqrt{mn}))=0$

$\Rightarrow mx(mx+n+\sqrt{mn})+(n-\sqrt{mn})(mx+n+\sqrt{mn})=0$

$\Rightarrow (mx+n+\sqrt{mn})(mx+n-\sqrt{mn})=0$

Now, one of the products must be equal to zero for the whole product to be zero.

Hence, we equate both the products to zero in order to find the value of $\text{'}x\text{'}$

$(mx+n+\sqrt{mn})=0$

$\Rightarrow mx=-n-\sqrt{mn}$

$\Rightarrow x=\frac{-n-\sqrt{mn}}{m}$

(or)

$(mx+n-\sqrt{mn})=0$

$mx=-n+\sqrt{mn}$

$x=\frac{-n+\sqrt{mn}}{m}$

The roots of the above mentioned quadratic equation are

$x=\frac{-n+\sqrt{mn}}{m}\text{\hspace{0.17em}and\hspace{0.17em}}x=\frac{-n-\sqrt{mn}}{m}$