If by eliminating $x$ between the equation $x^2 + ax + b = 0$ and $xy+
l(x + y) + m = 0$, a quadratic in $y$ is formed whose roots are the
same as those of the original quadratic in $x$. Then prove either $a =
2l$ & $b = m$ or $b + m = al$.
Initially, I thought of eliminating $x$ but that method seems to be very lengthy. Is there any trick that can help to solve the problem faster?
Best Answer
If $(x_1,y_1), (x_2,y_2)$ are the two solutions, then $y_1, y_2$ are the two solutions of the quadratic in $y$
We have two cases:
1) $x_1 = y_1, x_2=y_2$
This case $x=y$ and the second equation becomes $x^2 + 2lx + m=0$ therefore $2l = a, m=b$
2) $x_1 = y_2, x_2=y_1$
We know $x_1y_1+ l(x_1 + y_1) + m = 0$ Replacing $y_1$ with $x_2$ we get $b -al +m =0$