If by eliminating x between the equations x² +ax+b=0 and xy+l(x+y)+m=0, a quadratic equation in y is formed whose roots are the same as those original quadratic in x, the

Eliminating x we have
$\begin{array}{r}{\left(\frac{\mathrm{m}+\mathrm{ly}}{\mathrm{y}+\mathrm{l}}\right)}^2-\mathrm{a}\left(\frac{\mathrm{m}+\mathrm{ly}}{\mathrm{y}+\mathrm{l}}\right)+\mathrm{b}=0\text{ or }\\ (\mathrm{m}+\mathrm{ly}{)}^2-\mathrm{a}(\mathrm{m}+\mathrm{ly}\left)\right(\mathrm{y}+\mathrm{l})+\mathrm{b}(\mathrm{y}+\mathrm{l}{)}^2=0\end{array}$
is same as first eq 
$\therefore \frac{{\mathrm{l}}^2-\mathrm{al}+\mathrm{b}}{\mathrm{l}}=\frac{2\mathrm{lm}-{\mathrm{al}}^2-\mathrm{am}+2\mathrm{bl}}{\mathrm{a}}=\frac{{\mathrm{m}}^2-\mathrm{alm}+{\mathrm{bl}}^2}{\mathrm{b}}$

From 1st and third fraction $\mathrm{b}\left({\mathrm{l}}^2-\mathrm{al}+\mathrm{b}\right)={\mathrm{m}}^2-\mathrm{alm}+{\mathrm{bl}}^2\text{ or }(\mathrm{b}-\mathrm{m})(\mathrm{al}-\mathrm{b}-\mathrm{m})=0$

either b = m or b + m = al From 1st and second fraction, putting b = m
$\begin{array}{l}{\mathrm{al}}^2-{\mathrm{a}}^2\mathrm{l}+\mathrm{am}=4\mathrm{lm}-{\mathrm{al}}^2-\mathrm{am}\text{ or }\\ 2{\mathrm{al}}^2-{\mathrm{a}}^2\mathrm{l}-4\mathrm{lm}-2\mathrm{am}=0\text{ or }\\ {\mathrm{a}}^2\mathrm{l}-2\mathrm{a}\left({\mathrm{l}}^2+\mathrm{m}\right)+4\mathrm{lm}=0\therefore \mathrm{a}=2\mathrm{l}\text{ or }\mathrm{al}=2\mathrm{m}\end{array}$

Thus either b = m and a = 2l    b = m   and   al = 2m