Question

Cuales son las raices de la ecuacion

mnx^2+mn=x(m^2+n^2)

Answer 1

Solucion

formamos una ecuación de la forma  ax²+bx+c =0
para aplicar la formula general

$a=mn \\ b=-( m^{2} + n^{2}) \\ c= mn \\ \\ mn x^{2} -x( m^{2} + n^{2})+mn= \\ formula \\ \\ x=-b+- \frac{ \sqrt{ b^{2} -4ac} }{2a} =remplazamos \\ \\ \frac{-.[-( m^{2} + n^{2})]+- \sqrt{-[-( m^{2} + n^{2})-4mn.mn } }{2mn} = \\ \\ \\ x = \frac{ m^{2} + n^{2} +- \sqrt{ [-( m^{2} + n^{2} )^{2} -4 m^{2} n^{2} } }{2mn} \\ \\ x= \frac{ m^{2} + n^{2}+- \sqrt{ m^{4}+2 m^{2} n^{2} + n^{4} -4 m^{2} n^{2} } }{2mn} \\ \\ [tex] \\ \\ x= \frac{ m^{2} + n^{2} +- \sqrt{ m^{4}-2 m^{2} n^{2} + n^{4} } }{2mn} \\ \\ \\ \\ x= \frac{ m^{2} + n^{2} +- \sqrt{ ( m^{2} - n^{2}) ^{2} } }{2mn} = \\ \\ x= \frac{ m^{2} + n^{2}+-( m^{2} - n^{2} ) }{2mn} = \\ \\ \frac{ m^{2} + n^{2} + m^{2} - n^{2} }{2mn} = \\ \\ x= \frac{2 m^{2} }{2mn} = \frac{m}{n} ..primera...respuesta \\ \\ x= \frac{ m^{2} + n^{2}- m^{2} + n^{2} }{2mn} = \\ \\ x= \frac{2 n^{2} }{2mn} = \frac{n}{m} ...segunda respuesta \\ \\ saludos$