m/nx² +n/m =1-2x. Solve for x. …

According to the question,
{tex}\frac{m}{n}{x^2} + \frac{n}{m} = 1 - 2x{/tex}
{tex}\Rightarrow \frac{m}{n}{x^2} + 2x + \frac{n}{m} - 1 = 0{/tex}
{tex} \Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2}}}{{{m^2}}} - \frac{n}{m} = 0{/tex} [multiplying both sides by 'n' and dividing both sides by 'm']
{tex}\Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
To factorize {tex}{x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}}{/tex}, we have to find two numbers {tex}'a'\ and\ 'b'{/tex} such that.
{tex}a + b = \frac{{2n}}{m}{/tex} and {tex}a b = \frac { n ^ { 2 } - m n } { m ^ { 2 } }{/tex}
Clearly, {tex}\frac{{n + \sqrt {mn} }}{m} + \frac{{n - \sqrt {mn} }}{m} = \frac{{2n}}{m}{/tex} and {tex}\frac{{\left( {n + \sqrt {mn} } \right)}}{m}\times\frac{{\left( {n - \sqrt {mn} } \right)}}{m} = \frac{{{n^2} - mn}}{{{m^2}}}{/tex}  ({tex}\therefore a = \frac{{n + \sqrt {mn} }}{m}{/tex} and {tex}b = \frac{{n - \sqrt {mn} }}{m}{/tex})
{tex}\Rightarrow {x^2} + \frac{{2nx}}{m} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} + \frac{{\left( {n + \sqrt {mn} } \right)}}{m}x + \frac{{\left( {n - \sqrt {mn} } \right)}}{m}x{/tex}{tex} + \frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}
{tex} \Rightarrow x\left[ {x + \frac{{n + \sqrt {mn} }}{m}} \right] + \frac{{n - \sqrt {mn} }}{m}{/tex} {tex}\left[ {x + \frac{{n + \sqrt {mn} }}{m}} \right] = 0{/tex}
{tex} \Rightarrow \left( {x + \frac{{n - \sqrt {mn} }}{m}} \right)\left( {x + \frac{{n + \sqrt {mn} }}{m}} \right) = 0{/tex}
{tex} \Rightarrow x + \frac{{n - \sqrt {mn} }}{m} = 0{/tex} or {tex}x + \frac{{n + \sqrt {mn} }}{m} = 0{/tex}
{tex} \Rightarrow x = \frac{{- n - \sqrt {mn} }}{m}{/tex} or {tex}x = \frac{{ - n + \sqrt {mn} }}{m}{/tex}