Quadratic trouble#2

Since we have one of the roots double the other let's call them r, 2r we can transform the equation by divding (l-m) \not necessary step

$\left( l-m \right) { x }^{ 2 }+lx+1=0\quad \Longrightarrow \quad { x }^{ 2 }+\frac { l }{ (l-m) } x+\frac { 1 }{ (l-m) } =0$

and then we can conclude that the product of the two roots is

$2{ r }^{ 2 }=\frac { 1 }{ (l-m) }$

and then that their sum is

$3r=\frac { l }{ (l-m) }$

by substituting in the first equation and simplifying we get

$2\frac { { l }^{ 2 } }{ 9{ (l-m) }^{ 2 } } =\frac { 1 }{ (l-m) } \\ 2{ l }^{ 2 }=9(l-m)\\ 2{ l }^{ 2 }-9l+9m=0$

we can then use the quadratic formula to solve for L and since we know that L is not a complex number since one of the roots is real so we should have the determinant greater than or equal zero, we can then find the greatest value for m and you could substitute it back and solve for L and then substitute both in the equation to test if they lead to meaningful results.

$\frac { 9\pm \sqrt { 81-72m } }{ 4 }$

$81-72m\ge 0$

$m\le \frac { 9 }{ 8 }$