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Problem

Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.

Source

This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time solving it, so thought I'd better ask here.

Observations

  • Some real roots are possible: when $a<0$, the equation has two of them.
  • If one more coefficient was allowed to be arbitrary: $a x^4 + b x^3 + cx^2 + x + 1 = 0$, then the roots could be all real, since every quartic can be brought into such form by scaling
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  • $\begingroup$ If $x=1$ is not a root you can apply Rolle's theorem to $f(x)(x-1)$. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 17:47
  • $\begingroup$ But what will that give me. $\endgroup$ – user2369284 Nov 23 '13 at 17:51
  • $\begingroup$ But please share some of your thoughts and give a bit of background. Where does this problem come from? $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 17:51
  • $\begingroup$ Rolle's theorem implies that if an evereywhere differentiable function has $n$ distinct zeros, then its derivative has at least $(n-1)$ distinct zeros. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 17:53
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    $\begingroup$ @TitoPiezasIII: I think the missing context is some effort on OP's part. $\endgroup$ – tomasz Nov 23 '13 at 21:36
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Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But then the same is true of $f(x-\frac14) = x^4 + \frac58 x^2 + Bx + A$ for some $B$ and $A$ (we don't need the formula). If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$ then $a^2+b^2+c^2+d^2 = -2\frac58 < 0$, so $a,b,c,d$ cannot all be real, QED.

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    $\begingroup$ If I may add something? :) In general, if $f(x) = x^4+px^3+qx^2+rx+s=0$, then the sum of the square of the roots $x_i$ is $S(p,q)=x_1^2+x_2^2+x_3^2+x_4^2 = p^2-2q$. Hence $S(1,1)=-1$. Since the sum of the square of real numbers can't be negative, this implies not all the $x_i$ are real. $\endgroup$ – Tito Piezas III Nov 23 '13 at 20:28
  • $\begingroup$ Noam, I would like to ask , how did you choose f(x). Is this a set procedure for this kind of problems or you chose it randomly. $\endgroup$ – user2369284 Nov 24 '13 at 10:47
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    $\begingroup$ @user2369284 See Reciprocal polynomial. $\endgroup$ – user147263 Aug 6 '14 at 6:41
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Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But $f''(x) = 12x^2 + 6x + 2 > 0$ for all $x$. Therefore $f$ is convex upwards and hence can have at most two real roots, QED.

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Assume contrariwise that $f(x)=ax^4+bx^3+x^2+x+1$ has 4 distinct real zeros for some choice of $a,b$. Then $$ g(x)=f(x)(x-1)=ax^5+(b-a)x^4+(1-b)x^3-1 $$ has either five distinct zeros, or a double zero at $x=1$ together with three other zeros.

But $$ g'(x)=5ax^4+4(b-a)x^3+3(1-b)x^2 $$ has a double zero at $x=0$, and at most two other zeros.

Rolle's theorem tells us that the derivative of a differentiable function has a zero between any two of its zeros. This rules out the possibility of $g(x)$ having five distinct zeros.

The remaining possibility is that $g(x)$ has a double zero at $x=1$ and three other zeros. This means that $x=1$ is one of the zeros of $g'(x)$, so by Rolle's theorem $g'(x)$ should have three zeros different from $x=1$. We saw that this cannot be the case, so we have arrived at a contradiction.


How to see this? The fixed part of the three lowest degree terms of the polynomial $f(x)$ is known to be a factor of $x^3-1$. With nothing else to go by, I decided to check out, where using that bit takes us.

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  • $\begingroup$ I needed Rolle here, sorry :-( $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 18:28
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    $\begingroup$ It can also be a possibility that $g(x)$ has double zero at anywhere else other than 1. Also this only rules out that there can be 4 "distinct" real roots, not 4 real roots $\endgroup$ – user2369284 Nov 23 '13 at 18:45
  • $\begingroup$ @user2369284: Today I had reason to revisit this question. I see your objection now - somewhat belatedly. Very sorry about that. The argument becomes messier in those case. For example, if $f$ had real zeros $x_0<x_1<x_2<1$ of which $x_1$ is a double root, then $g'(x)$ should have a zero at $x_1$, and other zeros in the open intervals $(x_0,x_1)$, $(x_1,x_2)$ and $(x_2,1)$ which is again a contradiction. I haven't checked all the possibilities, but it seems to me that we always run into a contradiction with Rolle. $\endgroup$ – Jyrki Lahtonen May 6 '15 at 20:47

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