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| 1. |
m/n x2 + n/m = 1- 2x, solve by factorization method? |
| Answer» 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} | |