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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.

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    $\begingroup$ Shouldn't it be that ...prove that either "$a=2l$ and $b=m$" or "$b+m=al$"? $\endgroup$
    – mathlove
    Nov 15, 2016 at 16:59
  • $\begingroup$ @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. $\endgroup$
    – user350331
    Nov 15, 2016 at 19:14

1 Answer 1

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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 $al - m = b \iff b+m = al$.

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