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If by eleminating $x$ between the equation $x² + ax + b = 0$ & $xy + l (x + y) + m = 0$, 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 $b + m = al$.

Today while we were solving this sum in class our professor told us that rather than directly eliminating $x$ from both the equations there is a shortcut method too.Since the equation after eliminating $x$ in $y$ is same as the the original quadratic in $x$ we can simply replace $y$ by $x$ in the second equation to form another quadratic. Then we can apply the condition for one root common or both roots common between this new quadratic and the original quadratic to reach final result.

But I didn't get why we can replace $y$ by $x$ in the second equation.Can someone please explain in detail?Thanks.

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marked as duplicate by Alex M., Claude Leibovici, Frits Veerman, Mathematician 42, Alex Provost Apr 12 '17 at 17:44

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You can replace y in second quadratic equation because these two equation share the same roots: to make things clear, here is my solution:

$x² + ax + b = 0$ must have two roots: $x_1$,$x_2$.

For $xy + l (x + y) + m = 0$
if $x = x_1$, $y = x_2$, then by Vieta's theorem: $xy = b,x+y = -a \implies b+m=al \\\tag1$

if $x = x_1$, $y = x_1 $, then $x_1^2+2lx_1+m =0$, easy to show that $2l = a, m =b \\\tag2$

Annotation:

Why the step(1) and step(2) is valid is because $x$ firstly satisfy the quadratic equation$x² + ax + b = 0$ and also satisfy the $xy + l (x + y) + m = 0$. WLOG, let $x= x_1$, because if $x = x_2$, the step is same.

Since y in $xy + l (x + y) + m = 0$ has same value for $x$ in $x² + ax + b = 0$, the range of $y$ has two value (i.e. {$x_1$,$x_2$}). So we can divide it into two cases(i.e. step (1) or step(2) )

Finally, Since step (1) and step (2) is two different case: the answer should be : $ b + m = al$ or ($a = 2l$ and $b = m$). The parenthese represents to pack up the case.

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