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Show that the polynomial $$P(x):=x^4-6x+6$$ has no real roots.


We need to solve this problem without using calculus. This is a problem from my son's olympiad textbook. Since the degree of the polynomial is $4$, our task seems difficult. I tried to factor the polynomial. For example, I was expecting an expression like $(x^2+1)(x^2+x+1)$. But unfortunately it didn't.

Rational root theorem obviously fails. Because, real roots don't exist.

I tried to rewrite the polynomial

$$x^4-6x+6=(x^2+ax+b)(x^2+cx+d)$$

So, it sufficies to show that $a^2-4b<0$ and $c^2-4d<0$.

But, I am not sure, is it good track or not.

Thanks for advance .

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    $\begingroup$ Have you tried drawing its graph? The roots of the derivative are easy to find, and then you just have to compute the value of $P$ at its two local minima. $\endgroup$
    – Plop
    Apr 25, 2023 at 21:06
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    $\begingroup$ No calculus @Plop $\endgroup$
    – Bruno B
    Apr 25, 2023 at 21:06
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    $\begingroup$ Which textbook? $\endgroup$ Apr 25, 2023 at 21:09
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    $\begingroup$ Yet another solution: $x^4-6x+6=(x-2)^2+(x-1)^2[(x+1)^2+1]$ $\endgroup$ Apr 25, 2023 at 21:13
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    $\begingroup$ Welcome to MSE! This is a nice display of an Olympiad question - wonderful start, +1! $\endgroup$
    – dan_fulea
    Apr 25, 2023 at 22:04

8 Answers 8

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Write $$x^4-6x+6=(x^2-1)^2+2\left(x-\frac32\right)^2+\frac12$$ and then it is clear that $x^4-6x+6>0$ for all $x\in\mathbb{R}$.

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    $\begingroup$ Nice! Just a remark for everyone else: a univariate nonnegative polynomial is always a sum of squares. This has been proved by Hilbert and the problem has been generalized and became famous (see en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem). I did not remember this fact when I saw the question yesterday, so this is why I'm commenting here. $\endgroup$
    – Plop
    Apr 26, 2023 at 10:17
  • $\begingroup$ But how did you find this polynomial? $\endgroup$ May 1, 2023 at 19:13
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    $\begingroup$ @KamalSaleh We wish to write $x^4-6x+6$ as a sum of squares so we can complete the square. If the first square is $(x^2+bx+c)^2$ to give us the $x^4$ term, we require $b=0$ (else there is a nonzero $x^3$ term), and $c$ must be small (if $c$ is too large, we will end up with a quadratic polynomial that can be negative added on to $(x^2+bx+c)^2$). So I tried $(x^2-1)^2$ and then completed the square for the quadratic that must be added to make the sum equal to $x^4-6x+6$. $\endgroup$
    – A. Goodier
    May 1, 2023 at 19:20
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Still another way :

If $x≤0$, then $x^4-6x≥0\thinspace .$ Therefore, $x>0\thinspace .$ Thus, using the AM-GM inequality you have :

$$x^3+\frac 2x+\frac 2x+\frac 2x=6≥4\sqrt [4]{8}$$

A contradiction .

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    $\begingroup$ Or more directly $x^4+2+2+2\geqslant 4\sqrt[4]8|x| \geqslant 6x$ with equality impossible simultaneously for both inequalities. +1 $\endgroup$
    – Macavity
    Apr 26, 2023 at 4:26
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We can use for example

$$x^4-6x+6 =(x^2-1)^2+2x^2-6x+5 >0$$

indeed $(x^2-1)^2\ge 0$ and

$$36-4\cdot 2\cdot 5 =-4<0$$

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Hint: $\,P(x+1) = x^4 + 4 x^3 + 6 x^2 - 2 x + 1 = x^2(x+2)^2 + x^2 + (x-1)^2\,$.

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  • $\begingroup$ You can actually rewrite it as $x^2(1+(x+2)^2)+(x-1)^2$, making it more obvious that the polynomial is greater than $0$. But I wonder how you found the second equality. $\endgroup$ May 1, 2023 at 19:14
  • $\begingroup$ @KamalSaleh I just completed the squares $x^4+4x^3+\color{blue}{4x^2}$ and $\color{blue}{x^2}-2x+1$ then what was left happened to be just $x^2$. $\endgroup$
    – dxiv
    May 1, 2023 at 19:41
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    $\begingroup$ @KamalSaleh I think it is obvious that the polynomial is never zero - a sum of squares can only be zero if all the squares are individually zero and the last two obviously can't be simultaneously zero. $\endgroup$ May 2, 2023 at 19:21
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This question attracted a huge echo, so let us write down a further decomposition of $f$ into a sum of squares: $$ x^4-6x+6=\left(x^2 -\frac 32\right)^2 + 3(x-1)^2 + \frac 34\ge \frac 34=0.75\ . $$ $\square$


Note: In fact, using analytic tools, we can immediately catch the sign of $f'(x)=4x^3-6$, it is negative on $(-\infty, a)$, positive on $(a,\infty)$, here $a=(3/2)^{1/3}\approx 1.1447142425533\dots$, so $f$ decreases on the first interval, increases on the second one. So $a$ is the point where the best lower bound, the global minimal value $f(a)\approx 0.8487859085100\dots$ is taken.

Above, we obtained the lower bound $0.75$ which is good enough for the purpose of excluding real roots.

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Another way, for $x<1$ we have $6-6x>0$ and for $x\ge1$

$$(x-1)^4+4(x-1)^3+6(x-a)^2+b=x^4+(8-12a)x+6a^2+b-3$$

which leads to $a=\frac76$ and $b=\frac 56$ that is

$$x^4-6x+6=(x-1)^4+4(x-1)^3+6\left(x-\frac76\right)^2+\frac56>0$$

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  • $\begingroup$ Instead I get: $(x-1)^4+4(x-1)^3+6(x-a)^2+b=x^4+(8-12a)x+6a^2+b\color{red}{-3}$. $\endgroup$ May 3, 2023 at 9:11
  • $\begingroup$ Thanks I check it! $\endgroup$
    – user
    May 3, 2023 at 9:23
  • $\begingroup$ Are you sure that the argument is waterproof? $(x-1)^4+4(x-1)^3=(x-1)^3(x+3)$ is negative for $x=0$. $\endgroup$ May 3, 2023 at 17:43
  • $\begingroup$ Yes of course! We need to distinguish the two cases. For $x<1$ we have $6-6x>0$ and then for $x\ge1$ we can use the given decomposition. $\endgroup$
    – user
    May 3, 2023 at 18:02
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To pursue your approach of factoring the quartic is a little troublesome because the exact expressions for the coefficients prove to be pretty terrifying (we catch the "bad end" of Ferrari's method).

We can conclude from the Rule of Signs that $ \ x^4 - 6x + 6 \ $ has either two or no positive and no negative real zeroes.* If we posit that there are two real and two complex-conjugate zeroes, we can begin with two quadratic factors $ \ (x^2 - ax + b)·(x^2 - 2cx + 6/b) \ \ , \ $ with $ \ c \ $ being the real part of the complex zeroes and $ \ \frac{6}{b} \ $ being the square of their modulus. The cubic term gives us $ \ a + 2c \ = \ 0 \ \ , \ $ so using this leads to equations for the quadratic and linear coefficients $$ b \ + \ \frac{6}{b} \ - \ a^2 \ \ = \ \ 0 \ \ \ , \ \ \ ab \ - \ \frac{6a}{b} \ \ = \ \ -6 \ \ . $$

$ ^{*} $ We can also see that $ \ x^4 - 6x \ = \ x·(x^3 - 6) \ $ has a zero at $ \ x \ = \ 0 \ $ and a "triple zero" at $ \ x \ = \ \sqrt[3]{6} \ \ , \ $ so "raising" the function curve by $ \ 6 \ $ units would produce a transformed curve with at most two positive real zeroes, or none if the minimum is $ \ y \ > \ -6 \ \ . \ $ (It turns out that it is, but only very slightly; this can be should be an inequality argument without calculus, but it is not simpler than the other (easier-to-follow) inequality-based answers already presented.)

Solving the non-linear $ \ a^2 \ = \ b \ + \ \frac{6}{b} \ \ , \ \ a·\left( \frac{6}{b} \ - \ b \right) \ = \ -6 \ \ $ simultaneously is what produces the rather complicated exact values for $ \ a \ $ and $ \ b \ \ . \ $

With a bit of computational help, we obtain a pair of solutions $ \ a \ \approx \ -2.35046 \ \ , \ \ b \ \approx \ 1.4857 \ \ ; \ \ a \ \approx \ +2.35046 \ \ , \ \ b \ \approx \ 4.0385 \ \ , \ $ corresponding to the factorization $ \ x^4 - 6x + 6 \ = \ (x^2 - 2.35046·x + 1.4857)·(x^2 + 2.35046·x + 4.0385) \ \ . $ Since $ \ \large{\frac{2.35046^2}{4} } \ \normalsize{\approx \ 1.38104 \ \ ,} \ $ neither of these factors has a non-negative discriminant. Both factors are irreducible over the real numbers, so the quartic polynomial has no real zeroes. (From the remark about how the complex zeroes are manifested in the coefficients of the quadratic factors, we could now "read off" what the four zeroes of this polynomial are.)

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The discriminant of the depressed quartic $p(x)=x^4+dx+e$ is $\Delta=-27d^4+256e^3$. In our example, $d=-6$ and $e=6$, hence $\Delta=2^43^347>0.$ For quartics with real number coefficients, if the discriminant is positive, then the roots are either all real or all non-real. In this example, we can not have $4$ real roots due to Decartes's rule of signs. See also boojum's answer. We conclude that $P(x)>0$ for all $x\in\Bbb R$.

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