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प्रश्न
Solve the following quadratic equations by factorization:
`m/nx^2+n/m=1-2x`
उत्तर
We have been given
`m/nx^2+n/m=1-2x`
`(m^2x^2+n^2)/(mn)=1-2x`
m2x2 + 2mnx + (n2 - mn) = 0
`m^2x^2+mnx+mnx+[n^2-(sqrt(mn))^2]=0`
`m^2x^2+mnx+mnx+(n+sqrt(mn))(n-sqrt(nm))+(msqrt(mnx)-msqrt(mnx))=0`
`[m^2x^2 + mnx + msqrt(mnx)]+[mnx-msqrt(mnx)+(n+sqrt(mn))(n-sqrt(mn))]=0`
`[m^2x^2 + mnx + msqrt(mnx)]+[(mx)(n-sqrt(mn))+(n+sqrt(mn))(n-sqrt(mn))]=0`
`(mx)(mx+n+sqrt(mn))+(n-sqrt(mn))(mx+n+sqrt(mn))=0`
`(mx+n+sqrt(mn))(mx+n-sqrt(mn))=0`
Therefore,
`mx+n+sqrt(mn)=0`
`mx=-n-sqrt(mn)`
`x=(-n-sqrt(mn))/m`
or
`mx+n-sqrt(mn)=0`
`mx=-n+sqrt(mn)`
`x=(-n-sqrt(mn))/m`
Hence, `x=(-n-sqrt(mn))/m` or `x=(-n-sqrt(mn))/m`
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