Solve the following quadratic equations by factorization:

Solution:

We have been given

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

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

$m^{2} x^{2}+2 m n x+\left(n^{2}-m n\right)=0$

$m^{2} x^{2}+m n x+m n x+\left[n^{2}-(\sqrt{m n})^{2}\right]=0$

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

$\left[m^{2} x^{2}+m n x+m \sqrt{m n} x\right]+[m n x-m \sqrt{m n} x+(n+\sqrt{m n})(n-\sqrt{m n})]=0$

 

$\left[m^{2} x^{2}+m n x+m \sqrt{m n} x\right]+[(m x)(n-\sqrt{m n})+(n+\sqrt{m n})(n-\sqrt{m n})]=0$

$(m x)(m x+n+\sqrt{m n})+(n-\sqrt{m n})(m x+n+\sqrt{m n})=0$

 

$(m x+n+\sqrt{m n})(m x+n-\sqrt{m n})=0$

Therefore,

$m x+n+\sqrt{m n}=0$

$m x=-n-\sqrt{m n}$

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

or,

$m x+n-\sqrt{m n}=0$

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

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

Hence, $x=\frac{-n-\sqrt{m n}}{m}$ or $x=\frac{-n+\sqrt{m n}}{m}$.