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Knowing that lines $y=a_1 x+b_1 x^2$ and $y=a_2 x+b_2 x^2$ coincide in every point we can infer that $a_1=a_2$ and $b_1=b_2$. This inference is self-evident for me, but I don't know how to prove it formally.

How to prove this inference formally, using predicate logic? It would be nice to receive answer in terms of Fitch notation.

As far I understand $ \left\{ \begin{array}{c} y=a_1x+b_1 x^2 \\ y=a_2x+b_2 x^2 \end{array} \right. $ stands for crosspoint of two lines.

lone student
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Alex Alex
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    Hint: If they are coincident at every point, then they are coincident at three points $(u,v),(s,t),(m,n)$. – Trebor Jul 12 '21 at 09:46
  • $,y_1(x)-y_2(x),$ is a quadratic with $> 2$ roots, so it is the zero polynomial by the linked dupe. – Bill Dubuque Jul 12 '21 at 10:59
  • I disagree that my question is duplicate. I saw "Why can a quadratic equation have only 2 roots?" and I was stil unable to answer my question. Maybe mine and the other one question seems similar to mathematicians, but for me as chemist they are quite different. For me these questions have nothing in common. The only answer is given here didn't satisfy my either. – Alex Alex Sep 05 '21 at 12:57

1 Answers1

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If the $y_1=a_1x+b_1x^2$ and $y_2=a_2x+b_2x^2$ coincide in every point, this means $y_1=y_2,\forall x\in\mathbb R$.

Let $x≠0$, then we have

$$\begin{align}&a_1x+b_1x^2=a_2x+b_2x^2 \\ \implies &x^2(b_2-b_1)+x(a_2-a_1)=0\\ \implies &x\left(x(b_2-b_1)+(a_2-a_1)\right)=0 \\ \implies &x(b_2-b_1)+(a_2-a_1)=0,\\ \implies &x=\frac{a_1-a_2}{b_2-b_1},~\text{if}~b_1≠b_2 \end{align}$$

But, that means if $x≠0$, then the coincide is at only one point, which gives a contradiction. Therefore, we deduce that

$$b_1=b_2$$

This means, $$a_1=a_2.$$

lone student
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