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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 $K[x^2 – 5x + 6] = x^2 – 3x + 2$

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2 Answers 2

Firstly, the hypothesis stated at the beginning is wrong. The correct statement would be something of the form (restricted to the real number case)

Suppose $f$ and $g$ are quadratic polynomials. Then if $f$ and $g$ have the same real roots (so either two real roots or one repeated real root), there exists a constant $K \neq 0$ such that $f(x) = K g(x)$ for all $x$

Because of this, your proof is flawed from the beginning, because the statement isn't true (as your example shows). The mistake in your proof which allows you to draw this conclusion is when you write that $x^2 - (p+q)x + pq = 0$ and $x^2 - (q+r)x + qr = 0$. Here you have written down are a pair of equations to solve for $x$, so from this you cannot deduce anything about the two polynomials for all values of $x$.

To prove the actual result, I believe using the remainder theorem should be enough in order to do so (if you don't know what this is, then the wiki article may be of use: http://en.wikipedia.org/wiki/Polynomial_remainder_theorem)

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Using my counter-example again, f = (x-1)(x-2) and g = (x-2)(x-3). Then f and g have the same common root (2). What should that K be? –  Mick Jan 9 at 3:56
    
@Mick There isn't such a $K$, unless I'm incorrectly interpreting what you are saying - the two polynomials are proportional to each other only if all of their roots agree. –  Andrew D Jan 9 at 15:39
    
Good explanation. If all the roots agree, then the statement becomes very very obvious. –  Mick Jan 9 at 16:03
    
In another word, the condition of two quadratic polynomials having a common root is not sufficient to say these polynomials are proportional, right? After some back search, I finally found the original statement quoted in the answer to question #628732 (which is in fact shown in the ‘Related’ column of this post). Would you like to take a look at that and give me your comment on it? –  Mick Jan 10 at 8:17
    
@Mick What you say is correct, yes. In that question (math.stackexchange.com/questions/628732/…), we also know that $a,b,c$ are in a geometric progression, i.e. that $b=ar$ and $c=ar^2$ for some $r \neq 0, and the idea of that question is to give a sufficient condition for an arbitary quadratic and a special type of quadratic to share a common root. Can you see the difference? –  Andrew D Jan 10 at 12:17

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.

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