2
$\begingroup$

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 -

Let,

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

Thanks.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\gamma$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.