# Homework Help: Polynomials question

1. Mar 14, 2013

### VertexOperator

1. The problem statement, all variables and given/known data

Find the minimum possible value for $$a^{2}+b^{2}$$ where a and b are real such that the following equation has at least one real root.

2. Relevant equations

$$x^{4}+ax^{3}+bx^{2}+ax+1$$

3. The attempt at a solution

I tried to find the roots of the equation and then find a relationship between them but that didn't workout very well for me.

2. Mar 14, 2013

### Staff: Mentor

There are general equations for those roots, but I don't think they are useful here.

- For b=2, there is a nice factorization. This might correspond to the solution, but I don't know how to show that.
- For minimal a^2+b^2, the polynomial has exactly one real root of degree 2. As complex roots come in pairs (for real coefficients), the polynomial can be written as (x+c)^2(x^2+dx+e) for some real c,d,e. It might be possible to solve this directly.

3. Mar 14, 2013

### hapefish

Here is another way to think about it:

If $x^4+ax^3+bx^2+ax+1$ has at least one real root, then that means its graph must intersect the x-axis. Since this is a 4th degree equation with a positive leading coefficient the tails of the graph both go off towards positive infinity. Therefore, if the minimum point on the graph is negative, then the graph crosses the x-axis at some point, and therefore it has real roots.

This would change your problem statement to: "Find the minimum possible value of $a^2+b^2$ such that the minimum value of $x^4+ax^3+bx^2+ax+1$ is less than zero"

This will still be a mess, but perhaps it is a more manageable mess than the original? (I haven't tried to solve it, but if this is a multivariable calculus class perhaps you could turn this into a constrained optimization problem with variables a, b, and x???)

Last edited: Mar 14, 2013
4. Mar 14, 2013

### hapefish

I think this it:

You can show that any real solution of $x^4+ax^3+bx^2+ax+1=0$ is also a solution of $x^4-ax^3+bx^2-ax+1=0$. Then you can combine these two equations and get that this solution must be a solution of $x^4+bx^2+1=0$. But this is great, because you can show from here that $b^2 \geq 4$. Since you know that letting a=0 and b=2 gives you real solution, you can be confident that this also gives you the least solution.

5. Mar 14, 2013

### VertexOperator

How can we show that any real solution of $x^4+ax^3+bx^2+ax+1=0$ is also a solution of $x^4-ax^3+bx^2-ax+1=0$?

6. Mar 14, 2013

### epenguin

That is a 'reciprocal equation' - if x is a root, 1/x is also a root.

It can be solved conveniently dividing it by x2 and expressing it in the variable z = (x + 1/x).

You might see a way to answer the question on the way before getting to the solution of the equation.

Last edited: Mar 14, 2013
7. Mar 14, 2013

### Bacle2

Not quite:

$x^4+4x^3+6x^2+4x+1=(x+1)^3$ , but $x^4-4x^3+6x^2-4x+1=(x-1)^3$

(BTW, this gives you a not-so-great lower bound of $4^2+6^2=52$ for $a^2+b^2$ **)

Still, you do know that if there is one real root, there are at least two, since if

you have a real root, your polynomial splits as a product of a monic and a cubic

and every cubic ( or odd-degree poly.) has a real root, since complex roots come in pairs.

So you get $x^4+ax^3+bx^2+ax+1=(x^2+cx+d)(x^2+ex+f)$

where, WOLG the left poly $(x^2+cx+d)$ splits *

You can then play around with the Galois groups to make sure the lefthand poly. splits.

Note too, that if x<0 and b is positive, it makes it harder for your poly. to have a root, so you want it to be small if positve, or,

even better, to have b negative. Similarly, you want |a|>|b| , since , for x negative, the a coefficient lower the value of

f(x) and a raises it.

* If the poly. doesn't split, then you must acquit.

** Note that $x^4+x^3+x+1$ , which is the substitution a=1 and b=0 , has -1 as a double-root.

Last edited: Mar 14, 2013
8. Mar 14, 2013

### Dick

Oooh. That's useful. So, in the extreme case where there is exactly one real root, then x=1/x. Nice hint.

9. Mar 14, 2013

### Bacle2

You mean up to multiplicity, right? Otherwise , if there is one real root, there must be at

least two real roots ( up to multiplicity), since then f(x) splits as the product of a monic and

cubic .

And by epenquin's answer , the second poly. is of the form

$x^2+ex+1$ , and the first is $(x±1)^2$

And then it is relatively-clean to expand :

$(x^2+ex+1)(x±1)^2$ and set it equal to f(x) (which I'll do later when I have more time).

Last edited: Mar 14, 2013
10. Mar 14, 2013

### VertexOperator

How is it even possible that there is only 1 real root? Since it is a quartic shouldn't be there 0, 2, or 4?

11. Mar 14, 2013

### Bacle2

Yes, you're right, (I mentioned that in my post); if there is one root, then f(x) is the product

of a monic and a cubic , and the cubic must have at least one real root), so f(x) splits

as the product of 2 monics and a quadratic.

12. Mar 14, 2013

### VertexOperator

ok, but I thought we have to count multiplicities?

13. Mar 14, 2013

### Bacle2

But I think it makes a difference to check multiplicities, since there is a difference between a poly. with a double real root and a complex root (with its conjugate) , and having a poly. with four real roots.

Sorry, my posts were all-over the place; this problem is too difficult for me to do it on-and-off while doing something else.

Last edited: Mar 14, 2013
14. Mar 14, 2013

### Dick

You can count them if you want. But having at least one real root doesn't depend on the multiplicity. The idea I'm trying to follow is that as you decrease a^2+b^2 then at some point there is exactly one real root (and yes, it will have multiplicity at least 2 and possibly 4). That leads back to epenguins hint.

Last edited: Mar 14, 2013
15. Mar 14, 2013

### Dick

That's ok. Now you've got me worried. I'm trying to say that at the minimal value of a^2+b^2 there is exactly one real root. Hence that root satisfies 1/x=x. But suppose at the minimal value there are TWO real roots. Like 1/2 and 2 both multiplicity 2. Then as you decrease a^2+b^2 then both roots go away at the same time. That seems pretty unlikely, but it's still a gap that needs closing.

16. Mar 14, 2013

### Bacle2

Sorry, have not been too focused here, but maybe we can use the fact here that

1 is an upper bound for $a^2+b^2$ , since a=1 and b=0 leads to f(x) having -1

as a root.

17. Mar 14, 2013

### Dick

Sure. There's a lot of upper bounds. And 0 is certainly a lower bound where there are no real roots. The trick is to use some property that happens exactly at the lower bound. Keep thinking about it. If I assume there is exactly one real root at the lower bound with multiplicity 2, then I think I've got it. If there are two, I don't. It's a challenging question.

Last edited: Mar 14, 2013
18. Mar 15, 2013

### VertexOperator

This question is from the IMO btw.

19. Mar 15, 2013

### Staff: Mentor

Great. I was sure that the symmetry could be used in some way, but did not see the trick.

This simplifies my equation in post 2...

Edit: Hmm, that might be a problem.

Last edited: Mar 15, 2013
20. Mar 15, 2013

### voko

The equation is palindromic, so the standard technique is: $$x^4 + ax^3 + bx^2 + ax + 1 = 0 \\ \Rightarrow x^2 + ax + b + a/x + 1/x^2 = 0 \\ \Rightarrow X^2 + pX + q = 0$$ where $X = 1/x + x$. The resultant quadratic is then analyzed.

21. Mar 15, 2013

### Dick

Yeah, that was epenguin's hint. I was sort of hoping to get away with an argument that didn't require actually finding the roots.

22. Mar 15, 2013

### epenguin

I think if you think about it you find you can't get a real root in x if z is not real.

So the discriminant (a2 - 4b + 8) of the quadratic in z has to be positive. But whatever else that limits, it does not put any minimum on a2 nor b2 which can be as small as you like. Or rather a can be as small as you like and as long as b is small enough it is OK

So you only have to worry about the equation for x:

x2 - xz + 1 = 0

From that, or else just by looking at the function (x + 1/x) you realise that no z between -2 and 2 can correspond to real x. So at least one of the z has to be outside that range. After which it may not be very nice, but you have some handle on it.

Everything I say needs you check.

23. Mar 15, 2013

### epenguin

So was I. After the problem has been solved we might find the elegant argument we should have thought of in the first place.

Last edited: Mar 15, 2013
24. Mar 15, 2013

### Dick

Alright then. I think you can handle the two real zero case. If a root r isn't equal to +/-1, then the polynomial having two roots of multiplicity 2 is (x-r)^2*(x-1/r)^2. You can expand it out and find a and b. Then find extrema of a^2+b^2 by differentiating it. It's minimized at r=+/-1, but the value of a^2+b^2 is 52. Which is way too large, since Bacle found a case where it's 1. The other possibility is (x-1)^2*(x+1)^2. Which gives a^2+b^2=4. Still too big.

That blemish eliminated then we should be back on track. In the case of minimal a^2+b^2 there is one real root of multiplicity 2. Hence its value must be +1 or -1. The rest is easy.

Last edited: Mar 15, 2013
25. Mar 15, 2013

### epenguin

:yuck:

Oof, just as I predicted, having convinced myself by the argument I indicated of what the answer was, I looked at the equation and it is indeed obvious that the minimum (a2 + b2) for real roots is 0, since in that case it has 1 and -1 as roots.

No doubt the smartasses who compete in this despicable IMO competion are wised up to look for that sort of rubbish.

Last edited: Mar 16, 2013