If $a,b,c, a_1, b_1, c_1$ are rational and equations $ax^2+bx+c=0$ and $a_1x^2+b_1x+c_1=0$ have only one root in common then

Question

prove that $b^2-ac$ and $b_1^2-a_1c_1$ are perfect squares.

The only way I could work around this problem was assuming the roots would be rational, which would the discriminat would be a perfect square. Indeed, it is the right way to solve it, but the are the roots rational? I got the answer intuitively, but I would like a proper explanation to it.

Thanks!

The claim is false. Consider $x^2-3x+2=0$ and $x^2-4x+3=0$ where $(-3)^2-1\times 2=7$ and $(-4)^2-1\times 3=13$ are not perfect squares. — mathlove, Nov 23 '19 at 10:08

score 1 · Answer 1 · answered Nov 22 '19 at 14:36

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@Aditya you can begin by noting that if $x_0$ is the common root then one has

$$ ax_0^2 + b x_0 + c = a_1 x_0^2 + b_1 x_0 + c \iff ax_0^2 + b x_0 = a_1 x_0^2 + b_1 x_0 \iff ax_0 + b = a_1 x_0 + b_1 $$, where in the last equality we have considered that $x_0 \neq 0$.

answered Nov 22 '19 at 14:36

Marcelo Ng

374

So, looking at the above equality we have $x_0 = \frac{b_1 - b}{a - a_1} \in \mathbb{Q}$. Thus, the common root is rational. – Marcelo Ng Nov 22 '19 at 14:47
I am sorry, the second equation was written slightly wrong. It has been corrected now – Aditya Nov 22 '19 at 14:51

If $a,b,c, a_1, b_1, c_1$ are rational and equations $ax^2+bx+c=0$ and $a_1x^2+b_1x+c_1=0$ have only one root in common then

1 Answers1