Given an equation: $$ x^5 - 3x = 1 $$ Show that:

  • It has at least $1$ root on $(1, 2)$;
  • It has at least $3$ roots on $\Bbb R$

I've started with considering a function $f(x)$ for $x\in [1, 2]$: $$ f(x) = x^5 - 3x - 1 $$

Then calculating its value on the left and right sides of the closed interval yields: $$ f(1) = -3\\ f(2) = 25 $$

Applying the Intermediate Value Theorem yields that there exists a point for $x_0 \in [1, 2]$ such that $f(x_0) = 0$. Which means that indeed at least one root exists.

However, for the second part of the question if we consider $f(x)$ for $x \in \Bbb R$, the only way I see is to try and guess the intervals where the function changes its sign and then apply IVT again. Consider for example $f(x)$ for $x \in \{-2, -1, 1, 2\}$.

I see how derivatives could be to the rescue here, the problem is that I'm not allowed to use derivatives.

Is there a rigorous way to prove what's stated without guessing and without using derivatives? Thank you!

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    $\begingroup$ There is probably not an easy way to prove this, no. Certainly, none easier than finding intervals. If you know that the equation has no double-roots, then you only have to find two real roots (since an polynomial of odd degree without multiple roots has an odd number of real roots.) $\endgroup$ Jun 25, 2019 at 16:56
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    $\begingroup$ Invoking the Descartes Rule of Signs: the polynomial $f(x)=x^5-3x-1$ has coefficient sign sequence "$+--$"; one sign change means exactly one positive root. The sequence for $f(-x)$ is "$-+-$"; two sign changes means either $0$ or $2$ negative roots. If you can show there is a least one negative root (perhaps with an interval check), then you will have shown that there are exactly three real roots. $\endgroup$
    – Blue
    Jun 25, 2019 at 16:58
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    $\begingroup$ @Peter: Not clairvoyant; just a fan of the Rule of Signs. :) $\endgroup$
    – Blue
    Jun 25, 2019 at 17:02
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    $\begingroup$ @Blue that is a good place to start, thank you for the hint! $\endgroup$
    – roman
    Jun 25, 2019 at 17:03
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    $\begingroup$ As $f(x)\to-\infty$ as $x\to-\infty$, we are done as soon as we find one $x<1$ with $f(x)>0$. Tnanks to the big coefficient for $x$, it is clear that we need only check $x=-1$ $\endgroup$ Jun 25, 2019 at 17:07

6 Answers 6


Let us use Descartes' sign Rule

$f(x)=x^5-3x-1=0$, the number of sign changes in $f(x)$ is one and number of sigm changes in $f(-x)$ are two. So as per Descarte's rule this equation can have atmost one real positive and atmost two real negative roots. So Descarte's rules permits at most three real root which can be located as : $f(-\infty)<0, f(-1)=+1, f(0)=-1$ means there is at least one real root in $(-\infty, -1)$ and at least one in $(-1,0)$. Next $f(1)=-3, f(2)=+25$. So one real positive roots is in $(1,2)$. In all this equation has three real roots two negative and one positive.

  • $\begingroup$ Descartes' sign Rule only says that the number of negative roots is 2 or 0. Using $f(-1)=0$ involves guessing that -1 is a good point to evaluate the polynomial. The only way I know to fing the exact number of real roots of $f$ without using any guessing is Sturm's Theorem $\endgroup$
    – Helmut
    Jul 5, 2019 at 16:56
  • $\begingroup$ @Helmut 38, a correction, here the descarte's rule suggests ATMOST 2 real . negative roots.Thanks I will see Strum's Theorem , but I guess it involves derivatives.. $\endgroup$
    – Z Ahmed
    Jul 5, 2019 at 17:41

Graphical method can help. $$x^5-3x=1 \stackrel{x\ne 0}{\iff} x^4-3=\frac1x$$ Sketch:

$\hspace{4cm}$enter image description here

Note: You can quickly jump to the intervals of interest and compare the values and increase/decrease of the functions.

  • $\begingroup$ Of course computing a function table with hundreds of values will allow to pinpoint closely where sign changes happen. $\endgroup$ Jun 26, 2019 at 21:21
  • $\begingroup$ @LutzL, to sketch $x^4-3$ the intercepts are enough, $1/x$ is hyperbola. For $x>0$, the first increases and the second decreasee, so one intersection. For $x<0$, checking $x=-\sqrt[4]{3}$ is enough. $\endgroup$
    – farruhota
    Jun 27, 2019 at 4:41
  • $\begingroup$ Now try to apply the graphical method to $x^4-100=1/x$. Try to find both number of intersections and intervals of intersections with this and your method. $\endgroup$
    – farruhota
    Jun 27, 2019 at 4:47

(1)$f(x)=x^5-3x-1$ has one sign change in its coefficients so at most $1$ positive root. Also, $f(1).f(2)<0$ assures exactly one positive root lying in $(1,2)$.

(2) $f(-x)=-x^5+3x-1$ has two sign changes in its coefficients so at most $2$ negative roots OR more precisely $0$ or $2$ negative roots (possibility of exactly $1$ negative root is discarded due to pairwise occurrence of remaining complex conjugate roots). Now $f(-1).f(0)<0$ is sufficient to assure the existence of exactly $2$ negative roots of $f(x)$.


The coefficient $-3$ in the middle is dominant enough to allow to consider the two binomial expressions $x^5-3x$ and $-3x-1$ from the sides of the Newton polygon as separate approximations for large and small roots, with their real roots at $\pm\sqrt[4]3$ and $-\frac13$. To get a more precise connection to the given polynomial, consider the product of the binomials $$ (x^4-3)(x+\tfrac13)=x^5+\tfrac13x^4-3x-1. $$ This is close to the given polynomial $x^5-3x-1$ so that in moving from one to the other the roots do not change much. The pairwise distance between the $3$ real roots and the $2$ imaginary roots $\pm i\sqrt[4]3$ is large enough that the nature of these roots will not change under the perturbation to the given polynomial. To verify this more or less heuristic consideration, compute the values of the polynomial at enclosing intervals around $-\frac43, -\frac13,\frac43$, for instance $[-\frac32,-1]$, $[-\frac12,0]$, $[1,\frac32]$.

A bit more certainty and general applicability one gets from computing the root radius. Recall that for $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ a radius enclosing all the roots is obtained as $$ R=1+\max_{k<n}|a_k|\text{ or } R=\max(1,|a_{n-1}|+...+|a_1|+|a_0|) $$ which gives here $R=4$ in both cases. As there is a large degree gap, scaling the polynomial might help reduce the radius, set $x=2y$, then $$ f(2y)=32y^5-6y-1=32(y^5-\tfrac3{16}y-\tfrac1{32}) $$ which results in the bounds $R=2+\frac3{16}$ or $R=2$ for $x$. Now that the range of the roots is sufficiently restrained, computing values at $x=-2,-1,0,1,2$ is now a well-founded strategy. Counting sign changes as per Descartes then confirms that there are exactly two negative roots.

A more precise guessing can be performed using derivatives.

$f'(x)=5x^4-3$ has 2 real roots at $\pm\sqrt[4]{\frac35}$, thus $f(x)$ has local extrema there. Check the function value at the local maximum $-\sqrt[4]{\frac35}$, for simplicity check at the close-by point $x=-1$: $$ f(-1)=-1+3-1=1>0. $$ Together with $f(0)=-1$ and $f(-\infty)=-\infty$ one can conclude for the existence of $2$ negative roots.

There can be no more roots, as then by Rolle the derivative would have to have more than $2$ real roots.

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    $\begingroup$ OP says they can't use derivatives, and it's hard to see how you'd know to check the value of $f$ at $x = -1$ without having that knowledge. $\endgroup$ Jun 25, 2019 at 18:38
  • $\begingroup$ @MichaelSeifert I agree that you couldn't know that but checking at $x=-1, 0, 1$ is a standard easy way to start a problem like this. $\endgroup$ Jun 27, 2019 at 13:34

Quick and dirty without calculus. Plugging in some small integer values in $x^5 -3x -1$ (always a good place to start on this kind of problem) shows that the sign changes between $-2$ and $-1$, then again between $-1$ and $0$, then again before the values grow to $\infty$. That guarantees at least three real roots.

  • $\begingroup$ How exactly is this different from what OP already wrote in the question as working, but not completely satisfying? $\endgroup$ Jun 26, 2019 at 21:20
  • $\begingroup$ @LutzL I read your comment and then reread the question carefully, as I should have before. You;re right. Perhaps I have reassured the OP that this is in fact a good way to answer, and quite rigorous enough. $\endgroup$ Jun 26, 2019 at 21:27

This is inefficient and it may not always work but

$x^5 - 3x-1 = x(x^4-3)-1$.

If we can find $a<b$ but that $a(a^4 - 3) \ge 1$ and $b$ so that $b(b^4 -3) \le 1$, then we will done. (Why?[*])

If we can find an $a$ where $a < 0; k= a^4 - 3 < 0$ then $ak > 0$. And if $|a| \ge 1$ and $|k| \ge 1$ then $ak \ge 1$. And that will do for our $a$.

(Note: this is sufficient but much stronger and much more than necessary.)

That's easy enough. $k = a^4 - 3 \le -1$ implies $a^4 \le 2$ implies $|a| \le \sqrt[4]{2}$ so we just need $-\sqrt[4]{2} \le a \le -1$ for our $a$.

Likewise if we have $b >0$ but $j = b^4-3 < 0$. Because $bj < 0$ and $bj - 1< -1 < 0$. (And again sufficient, albeit much stronger than necessary.

So $b^4 -3 <0$ so $b^4 < 3$ so $|b| < \sqrt[4]{3}$ so any $b; 0 < b < \sqrt[4]{3}$ will do.

[*]Thus we have $\lim_{x\to -\infty} x^5 - 3x -1 = -\infty$ so there are plenty of $n < a$ so that $n^5 -3n -1 < 0$. And $a(a^4-3) -1 > 0$ so there is a root between $n$ and $a$. And $b(b^4-3) -1 < -2 < 0$ so there is a root between $a$ and $b$. And $\lim_{x\to \infty} x^5 - 3x -1 = \infty$ so there are plenty of $m> b$ so that $m^5 -3m -1 > 0$ so there is a root between $b$ and $m$.

But that was inefficient and required some educated guesses.


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