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Solve the following quadratic equation by factorization:
$\frac{m}{n}x^2+\frac{n}{m}=1-2x$
Given:
Given quadratic equation is $\frac{m}{n}x^2+\frac{n}{m}=1-2x$
To do:
We have to solve the given quadratic equation.
Solution:
$\frac{m}{n}x^2+\frac{n}{m}=1-2x$
Multiplying by $mn$ on both sides, we get,
$mn(\frac{m}{n}x^2+\frac{n}{m})=mn(1-2x)$
$m^2x^2+n^2=mn-2mnx$
$m^2x^2+2mnx+n^2-mn=0$
$ \begin{array}{l}
x=\frac{-( 2mn) \pm \sqrt{( 2mn)^{2} -4\left( m^{2}\right)\left( n^{2} -mn\right)}}{2\left( m^{2}\right)}\\
\\
x=\frac{-2mn\pm \sqrt{4m^{2} n^{2} -4m^{2} n^{2} +4m^{3} n}}{2m^{2}}\\
\\
x=\frac{-2mn\pm \sqrt{4m^{3} n}}{2m^{2}}\\
\\
x=\frac{-2mn\pm 2m\sqrt{mn}}{2m^{2}}\\
\\
x=\frac{-2m\left( n\pm \sqrt{mn}\right)}{2m^{2}}\\
\\
x=\frac{-\left( n\pm \sqrt{mn}\right)}{m}\\
\\
x=-\left( n-\sqrt{mn}\right) \ or\ x=-\left( n+\sqrt{mn}\right)\\
\\
x=\sqrt{mn} -n\ or\ x=-\left( n+\sqrt{mn}\right)
\end{array}$
The values of $x$ are $\sqrt{mn}-n$ and $-(n+\sqrt{mn})$.