Query about a statement on the consequence of two quadratic equations having a common root I have read an answer (in this site, but I lost the question number) saying something like the following:-
If the quadratic equations F(x) = 0 and f(x) = 0 have a common root, then the quadratics are proportional to each other. That is,
K[f(x)] = F(x); for some constant K.
I tried to ‘prove’ it because this is completely new to me.
For simplicity, we can assume that both equations are monic.
Let p, q be the roots of F(x) = 0 and q, r be the roots of f(x) = 0 such that q is their common root.
Then, $x^2 – (p + q) x + pq = 0$ and  $x^2 – (q + r) x + qr = 0$
Rewriting the above, we have 
$pq = –x^2 + (p + q) x$ ……………..(1)
$qr = –x^2 + (q + r) x$ ……………….(2)
[Added constraints:- p, q, r, x, x + (p + q), and x + (q + r) are not zero.]
If we let $\frac {p} {r} = K$, then dividing (1) by (2), we have
$\frac {–x^2 + (p + q) x} {–x^2 + (q + r) x} = \frac {p} {r} = K$
$K[–x^2 + (q + r) x] = [–x^2 + (p + q) x]$
$K[x^2 – (q + r) x] = [x^2 – (p + q) x]$
$K[x^2 – (q + r) x] + pq = [x^2 – (p + q) x] + pq$
$K[x^2 – (q + r) x] + (Kr)q = F(x)$
$∴ K[f(x)] = F(x)$
The proof seems to be nice. May be someone can point out what went wrong. This is because the ‘fact’ does not quite match with the following counter-example:-
1, and 2 are the roots of $x^2 – 3x + 2 = 0$
2, and 3 are the roots of $x^2 – 5x + 6 = 0$ such that 2 is the common root.
It seems that there is no K such that $K[x^2 – 5x + 6] = x^2 – 3x + 2$

If two monic polynomials $P$ and $Q$ of degree two with rational coefficients share a common root then that root is either rational or the two polynomials are equal.  This follows from the fact that the common root is also a root of the difference $P-Q$ of degree at most one.
