Question

`3/n x^2 n/m=1-2x` 

 

Answer 1

The given equation is 

`3/n x^2 n/m=1-2x` 

⇒`( 3^2x^2+n^2)/(mn)=1-2x` 

⇒`m^2x^2+n^2=mn-2mnx` 

⇒`m^2x^2+2mnx+n^2-nm=0` 

This equation is of the form `ax^2+bx+c=0` where `a=m^2,b=2mn`  and `c=n^2-mn`

∴  Discriminant, 

`D=b^2-4ac=(2mn)^2-4xxm^2xx(n^2-mn)=4m^2n^2-4m^3n^2=4m^3n>` 

So, the given equation has real roots.
Now, `sqrtD=sqrt4m^3n=2msqrtmn`  

∴` α=(-b+sqrt(D))/(2a)=(-2mn+2msqrtmn)/(2xxm^2)=(2mn(-n+sqrtmn))/(2m^2)=(-n+sqrtmn)/m`  

β=` α=(-b-sqrt(D))/(2a)=(-2mn-2msqrtmn)/(2xxm^2)=(2mn(-n-sqrtmn))/(2m^2)=(-n-sqrtmn)/m`  

Hence, `(-n+sqrtmn)/m` and `(-n-sqrtmn)/m`are the roots of the given equation.