Solve the following quadratic equations by factorization:
$\frac{m}{n} x^{2}+\frac{n}{m}=1-2 x$
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 $x=\frac{-n+\sqrt{m n}}{m}$.
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