if the roots of the equation l(m-n)x^2+m(n-l)x+n(l-m)=0 are equal.Show that m=2ln/l+n.

If the roots of the equation are equal, since the equation
is a quadratic, its discriminant has to be
zero.

Discriminant = delta = b^2 -
4ac

a,b,c, are the coefficients of the
equation:

delta = [m(n-l)]^2
-4l(m-n)*n(l-m)=

We'll expand the
square:

m^2*(n^2 - 2nl + l^2) - 4ln(ml - m^2 - nl + mn) =
0

m^2*n^2 - 2m^2*nl + m^2*l^2
- 4mnl^2 + 4lnm^2 + 4l^2*n^2 - 4lmn^2 =
0

m^2*n^2 - 2m^2*nl + m^2*l^2
= 4mnl^2 - 4lnm^2 - 4l^2*n^2 + 4lmn^2 (1)

We'll re-write
the
constraint:

m=2ln/l+n

m(l+n) =
2ln

We'll raise to square both
sides:

m^2*n^2 +2m^2*nl + m^2*l^2 =
4l^2*n^2

m^2*n^2 + m^2*l^2 = 4l^2*n^2 - 2m^2*nl
(2)

We'll substitute (2) in
(1):

4l^2*n^2 - 2m^2*nl  -
2m^2*nl = 4mnl^2 - 4lnm^2 - 4l^2*n^2
+ 4lmn^2

4l^2*n^2 - 4m^2*nl = 4mnl^2 - 4lnm^2 - 4l^2*n^2
+ 4lmn^2

We'll divide by
4:

l^2*n^2 = mnl^2 - lnm^2
+ lmn^2

2l^2*n^2 = mnl^2 +
lmn^2

We'll factorize by m*n*l to the right
side:

2l^2*n^2 = m*n*l(l +
n)

We'll divide by l*n:

2l*n =
m(l+n)

m = (2ln)/(l+n)
q.e.d