Question

Find the roots of the following equation, if they exist, by applying the quadratic formula:

`m/n x^2 + n/m = 1 - 2x`

Answer 1

The given equation is 

`m/n x^2 + n/m = 1 - 2x`

⇒ `(m^2x^2 + n^2)/(mn) = 1 - 2x` 

⇒ m²x² + n² = mn – 2mnx 

⇒ m²x² + 2mnx + n² – mn = 0 

This equation is of the form ax² + bx + c = 0 where a = m², b = 2mn and c = n² – mn.

∴  Discriminant, D = b² – 4ac

= (2mn)² – 4 × m² × (n² – mn)

= 4m²n² – 4m²n² + 4m³n²

= 4m³n > 0

So, the given equation has real roots.

Now, `sqrt(D) = sqrt(4m^3n) = 2msqrt(mn)`  

∴ `α = (-b + sqrt(D))/(2a)`

= `(-2mn + 2msqrt(mn))/(2 xx m^2)`

= `(2mn(-n + sqrt(mn)))/(2m^2)`

= `(-n+sqrt(mn))/m`  

`β = (-b- sqrt(D))/(2a)`

= `(-2mn-2msqrt(mn))/(2 xx m^2)`

= `(2mn(-n - sqrt(mn)))/(2m^2)`

= `(-n - sqrt(mn))/m`  

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