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 $-(n+\sqrt{mn})$.

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Updated on: 10-Oct-2022

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