So the questions says -

Let $f(x), g(x)$ and $h(x)$ be quadratic polynomials having positive leading coefficients and real and distinct roots. If each pair of them has a common root, then find roots of $f(x)+g(x)+h(x)=0$.

What I did -


$$ f(x) = a_1 (x-\alpha) (x-\beta), \\ g(x) = a_2 (x-\beta)(x-\gamma), \\ h(x) = a_3 (x-\gamma) (x-\alpha), \\ F(x):=f(x)+g(x)+h(x)$$ Now,

$$ F(\alpha) = a_2 (\alpha-\beta) (\alpha-\gamma) \\ F(\beta) = a_3 (\beta-\gamma) (\beta-\alpha) \\ F(\gamma) = a_1 (\gamma-\alpha) (\gamma-\beta)$$

I don't know how to proceed further. I referred to the solution, it just multiplies $F(\alpha), F(\beta) \text{, and } F(\gamma)$ and it comes out to be negative. And hence it concludes that roots of $F(x)=0$ are real and distinct. Can anyone explain why?


  • $\begingroup$ i would write $$a(x-x_1)(x-x_2)+b(x-x_3)(x-x_4)+c(x-x_5)(x-x_6)$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 29 '17 at 20:01
  • $\begingroup$ According to the question, each pair has a common root. So we can't write in your way. $\endgroup$ – Devansh Kumar Sep 29 '17 at 20:03
  • 1
    $\begingroup$ Claim: If $a,b,c$ are the three roots such that $f$ has roots $a,b$; $g$ has roots $a,c$; and $h$ has roots $b,c$, the roots of $F(x):=f(x)+g(x)+h(x)$ are given by $M\pm N$ where $M$ is the arithmetic mean of $a,b,c$ and $N=\frac 13\sqrt{P_2-e_2}$ with $P_2$ and $e_2$ denoting the 2nd power sum and 2nd elementary symmetric polynomial formed by $a,b,c$ respectively. $\endgroup$ – Prasun Biswas Sep 29 '17 at 20:15
  • $\begingroup$ A tedious way to prove the above claim would be to expand $$F(x):=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$$ and then apply the quadratic formula (note that $F$ having leading coefficient $1$ works w.l.o.g since we can always divide $F$ by its leading coefficient which doesn't change the roots. Though, I wonder if there's a more elegant proof for it. $\endgroup$ – Prasun Biswas Sep 29 '17 at 20:29
  • $\begingroup$ @PrasunBiswas You don't have, in general, all coefficients equal to 1. You can have their sum equal to 1. $\endgroup$ – Aretino Sep 29 '17 at 20:59

Suppose, without loss of generality, that $\alpha<\beta<\gamma$: it is easy to check that $F(\alpha)>0$, $F(\beta)<0$ and $F(\gamma)>0$. But $F(x)$ is a quadratic polynomial, hence a continuous function: it follows that $F(x)=0$ for some $x$ between $\alpha$ and $\beta$, and also for some $x$ between $\beta$ and $\gamma$.

  • $\begingroup$ I am unable to understand it this way. Can you please elaborate? $\endgroup$ – Devansh Kumar Sep 29 '17 at 20:28
  • $\begingroup$ If a continuous function $F(x)$ is such that $F(a)$ and $F(b)$ have different sign, then $F(x)$ vanishes for some $x$ between $a$ and $b$. It's called Bolzano's theorem. $\endgroup$ – Aretino Sep 29 '17 at 20:32
  • $\begingroup$ How do I check they are of different signs? And what if any two of them are equal? $\endgroup$ – Devansh Kumar Sep 29 '17 at 20:39
  • $\begingroup$ For instance: $F(\alpha)=a_2(α−β)(α−γ)$, but we know that $a_2>0$ ("positive leading coefficients"), and from $\alpha<\beta<\gamma$ we get $(α−β)<0$ and $(α−γ)<0$. Hence $F(\alpha)>0$, and you can repeat the argument to find $F(\beta)<0$ and $F(\gamma)>0$. $\endgroup$ – Aretino Sep 29 '17 at 20:43
  • $\begingroup$ The two roots of $F(x)$, let's call them $x_1$ and $x_2$, cannot be the same, because $\alpha<x_1<\beta$ and $\beta<x_2<\gamma$. $\endgroup$ – Aretino Sep 29 '17 at 20:45

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