# Quadratic Equation in $x$ and $y$

Problem Statement:-

If by eliminating $$x$$ between the equation $$x^2+ax+b=0$$ and $$xy+l(x+y)+m=0$$; a quadratic equation in $$y$$ is formed whose roots are same as those of the original quadratic equation in $$x$$, then prove that either $$a=2l$$, or $$b=m$$ or, $$b+m=al$$

Attempt at a solution:-

$$x^2+ax+b=0\tag{1}$$ $$xy+l(x+y)+m=0\tag{2}$$

As it is given that on eliminating $$x$$ from $$(1)$$ and $$(2)$$ we get a quadratic equation in $$y$$ which has the same roots as that of $$x$$ in equation $$(1)$$, so we can conclude that the solution for eq. $$(2)$$ is $$x=y$$.

Hence, eq. $$(2)$$ becomes $$x^2+2lx+m=0\tag{3}$$

As equations $$(1)$$ and $$(2)$$ have the same roots hence, we get $$\dfrac{1}{1}=\dfrac{a}{2l}=\dfrac{b}{m}$$

From this we get $$a=2l$$ and $$b=m$$.

I am not able to obtain the third relation, can you guide me in the right direction.

• Shouldn't it be that ...prove that either "$a=2l$ and $b=m$" or "$b+m=al$"? Nov 15, 2016 at 16:59
• @mathlove - Yeah I think so too, because the first two conditions kind of do come as a pair, hence were easy to figure out. But that's what the book has written so I left it as it is. Nov 15, 2016 at 19:14

As it is given that on eliminating $x$ from $(1)$ and $(2)$ we get a quadratic equation in $y$ which has the same roots as that of $x$ in equation $(1)$, so we can conclude that the solution for $(2)$ is $x=y$.

This is where you lost a set of solutions. It is true that the two equations have the same solutions, but $(1)$ is a quadratic and has two solutions. For $y$ to satisfy the same equation, it can be either $x$, or the other solution which is $(-a-x)$ by Vieta's formulas.

• The case $y=x$ was addressed in the original question, giving $a=2l$ and $b=m$.

• In the second case, substituting $y=-a-x$ in $(2)$ gives

$$x(-x-a) + l(x - x - a) + m = 0$$

$$-x^2 - a x - (al - m) = 0$$

Comparing coefficients to $(1)$ gives directly $al - m = b \iff b+m = al$.