if one root of equation (l-m)x2+lx+1=0 is double the other and lis real,then greatest value of m is

Hello Sakshi, given the zeros are p and 2p
Sum of the roots i.e p + 2p = 3p = - l /(​l-m) ---(1) and 
product of roots i.e 2 p^2 = 1/(l -m) --(2)
More over roots are real so discriminant > 0
So l^2 - 4 *(l -m) * 1 > 0
Or ​l^2 > 4 *(l -m) 
From (1) and (2) eleminating p we get ​l^2 = (4.5) *(l -m) 
So m must be less than ​l