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Problem: If $ax^2+bx+c=0$ and $2x^2 +3x+4=0$ have a common root where $a,b,c \in \Bbb N$, find least value of $a+b+c$

Solution: Here $2x^2 +3x+4=0$ will give complex roots

These roots will be in pair

Both equations have a common root

$ax^2+bx+c=0$ also have these complex pairs

This means least value will be obtained at $a=2$, $b=3$ and $c=4$

Least value of $a+b+c= 9$

Am I doing right?

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rst
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  • It is indeed true that the equations must have two common roots. But why does it mean $a=2, b=3, c=4$? –  Jan 13 '14 at 12:41
  • Yes it is correct. – GTX OC Jan 13 '14 at 12:41
  • @GTX: The last line is correct, but not all of the lines before it.... –  Jan 13 '14 at 12:42
  • @rst : It does not actually mean $a=2,b=3,c=4$ but the least value would be attained at that point and you are correct otherwise! –  Jan 13 '14 at 12:42
  • @PraphullaKoushik, thanks I got it. – rst Jan 13 '14 at 12:45
  • There is a few arguments missing: I would wrote "$ax^2+bx+c=0$ also have this complex pair (since it has real coefficients)".

    More importantly, two quadratic equations $ax^2+bx+c=0$ and $a'x^2+b'x+c'=0$ have the same solutions if and only if $a'=ka$, $b'=kb$ and $c'=kc$ for some nonzero real number $k$. Here, it is important that $a$, $b$ and $c$ are integers to conclude that the minimum of $a+b+c$ exists and is obtained for $a=2$, $b=3$ and $c=4$ (if $a$,$b$,and $c$ where rational numbers, then the minimum does not exist).

     – Taladris Jan 13 '14 at 12:52

2 Answers2

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What about actually finding the roots? After all this is high school stuff:

$$2x^2+3x+4=0\iff x_{1,2}=\frac{-3\pm\sqrt{23}\,i}4$$

and thus

$$ax^2+bx+c=a(x-x_1)(x-x_2)$$

By Vieta's formulas:

$$b=-(x_1+x_2)a=\frac32a\implies a\;\;\text{is even positive}$$

$$c=ax_1x_2=2a\implies c\;\;\text{is at least}\;\;4\;,\;\text{by the above line}$$

Well, there you go...

Martin Sleziak
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DonAntonio
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  • I haven't seen that formula referred as Viete's formula before. – Babai Jun 30 '16 at 07:09
  • @Babai Well, now you have. Check for example here: http://mathworld.wolfram.com/VietasFormulas.html , or also here: https://www.artofproblemsolving.com/wiki/index.php?title=Vieta%27s_Formulas – DonAntonio Jun 30 '16 at 07:11
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    @Babai My pleasure. We all learn new things here. This is one of the most important, if not the most important, features of this site. – DonAntonio Jun 30 '16 at 07:16
  • Absolutely true! – Babai Jun 30 '16 at 07:18
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Big gap! You implicitly assume that both equations will share two roots. This is the crux of the matter and it requires rigorous proof! (the accepted answer has the same gap). Here is one way.

Hint $ $ Call them $\,f(x)\,$ and $\,g(x).\,$ A common root is a also a common root of the polynomial $\,h\,$ obtained by eliminating their quadratic terms: $\,h = f-(a/2) g.\,$ Since the discriminant of $\,g\,$ is negative, both roots of $g$ are non-rational. So the only way either can be a root of the polynomial $\,h\,$ of degree $\le 1\,$ is if $\,h = 0.\,$ Thus $\,f = (a/2) g\,$ is a constant multiple of $\,g.\,$ The rest is straightforward.

Remark $\ $ More generally any common root of $\,f,g\,$ is also a root of every one of their linear combinations $\, h_1 f + h_2 g.\,$ But, by Bezout, we know that $\,\gcd(f,g)\,$ has that form, hence $\,f(a) = 0 = g(a)\,\Rightarrow\, \gcd(f,g)(a) = 0.$

Bill Dubuque
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  • "Since the discriminant of g is negative, both roots of g are irrational." Discriminant negative implies roots are Complex Number. (Irrational are real numbers). In your remark, what do you mean by linear combination? Do you mean $h_1$ and $h_2$ are constants? Your claim "More generally any common root of $f$,$g$ is also a root of every one of their linear combinations $h_1 f+h_2 g$." is not true . Take for example $x^2-2x+4$ and $x^2-x-2$ and $h_1=1$ and $h_2=-1$. – Babai Jun 30 '16 at 07:06
  • @Babai $(1)$ Like many algebraists, I use the general definition irrational = not rational, so complex numbers $\not\in \Bbb Q$ are irrational. $(2)$ Your $f$ and $g$ have no common root so they certainly satisfy my claim (e.g. every root of $f$ is a root of $fg$ is true (vacuously) even when $f$ has no roots in the ambient ring). Usually it is a good idea to ask for clarification before downvoting to help avoid misunderstandings like this. Anyway, to avoid confusion, I changed "irrational" to "non-rational". The edit enables downvote removal. – Bill Dubuque Jun 30 '16 at 13:40
  • My apologies! I understand. – Babai Jun 30 '16 at 13:50
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    The OP doesn't "implicitly assume" that both equations share two roots, they sketch an argument for that. The "complex" in "will give complex roots" should of course be "non-real", but it's common enough for people to use "complex" in that meaning. The sketch is not particularly well-formulated, but it's clear enough to see that the OP knows the reason why both equations share two roots. – Daniel Fischer Jun 30 '16 at 14:02
  • @Babai No problem, I can understand how one might be misled on first encountering that generalized definition (which might color subsequent judgements of the author). While Gelfond, Manin, Ribenboim, Shafarevich... can get away with that, alas, apparently I cannot! – Bill Dubuque Jun 30 '16 at 14:03
  • @DanielFischer I see no such proof in the OP, only an (unjustified) claim. – Bill Dubuque Jun 30 '16 at 14:06
  • A sketch of an argument is something different from a proof. But it often is sufficient a) to indicate that the author of the sketch knows how to fill in the missing parts to give a complete proof, and b) to enable the reader to fill in the missing parts themselves. – Daniel Fischer Jun 30 '16 at 14:10
  • @DanielFischer Please explain what argument you think is sketched. I don't see it. – Bill Dubuque Jun 30 '16 at 14:11
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    The two zeros of $2x^2 + 3x + 4$ are non-real and conjugate. Let $\zeta$ be a common zero of $2x^2 + 3x + 4$ and $ax^2 + bx + c$. Since $a,b,c \in \mathbb{R}$, it follows that $\overline{\zeta}$, the other zero of $2x^2 + 3x + 4$, is also a zero of $ax^2 + bx + c$, so both zeros of $2x^2 + 3x + 4$ are common zeros. – Daniel Fischer Jun 30 '16 at 14:14
  • @DanielFischer Possibly that is what the OP intended, but there is no way to be sure from what very little was written (e.g. no mention of conjugation). I've encountered many students who think that this is "obvious" so does not require proof. But when asked to prove it they could not do so. Proofs at this level should not leave such ambiguity in the mind of the reader. Hence my critique. – Bill Dubuque Jun 30 '16 at 14:28