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If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$

Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$ $$3bc-9ac-5b^2+15ab=c^2+25a^2-10ac$$ $$25a^2+5b^2+c^2=15ab+3bc+ac$$ There is pattern in the equation. The coefficients of the middle terms and first terms of both sides are $(5\times 5,5)$ and $(5\times 3,3)$.

I tried to use Lagrange multipliers. But isnt there any simpler way to minimize $a+b+c$?

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  • $\begingroup$ I guess the true problem should be with undetermined coefficients instead of $3$ and $5$. $\endgroup$
    – Piquito
    Commented Jan 28, 2016 at 14:35

3 Answers 3

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Hint:

The roots of $x^2+3x+5$ are not real, and if $z\in \Bbb C$ is a root of a polynomial with real coefficients but $z$ is not real, then $\bar z$ is also a root of this polynimial.

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The polynomial $x^2+3x+5$ is irreducible over $\mathbb{Q}$. Hence, if $x^2+3x+5$ and $ax^2+bx+c$, with $a,b,c\in\mathbb{Q}$, have a common root, they must be proportional. That is, $$ax^2+bx+c=a\left(x^2+3x+5\right)\,.$$

The problem would be more challenging if $x^2+3x+5$ is replaced by $x^2+3x+2$.

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See one root is complex so other os obviously its conjugate as $b^2-4ac<0$ so the equations have both roots in common and as $a,b,c$ have no common factor thus $min(a+b+c)=1+3+5=9$

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  • $\begingroup$ From where did you get 1+3+5 ? $\endgroup$
    – Shailesh
    Commented Feb 4, 2016 at 2:48
  • $\begingroup$ From the given equation $\endgroup$
    – Archis Welankar
    Commented Feb 4, 2016 at 4:56

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