# What are the conditions on $a$ such that the polynomial $x^4-2ax^2+x+a^2-a$ has four distinct real roots?

I was assisting with a local competition for high school students which was being run today, and one of the questions on the question paper was:

What are the conditions on the real number $a$ such that the polynomial $$x^4-2ax^2+x+a^2-a$$ has four distinct real roots.

The questions were all multiple choice, so it was possible to answer the question by trying each of the options and eliminating those which do not work. (e.g. By finding a value of $a$ satisfying the conditions specified in the option, but for which the polynomial does not have four distinct real roots)

The conditions on $a$ turn out to be that $a>\frac{3}{4}$, but I found this answer by using the approach above (and later by plotting the polynomial for various values of $a$, but this of course also does not prove anything). This is unsatisfying for two reasons: firstly that it is of course desirable to have a proof that this is the case, and also because one of the options given was "None", which could be taken to mean "There are no conditions on $a$", but which could also be interpreted to mean "None of the above", which would make the approach above invalid.

So my question is: how can we prove that the polynomial has four distinct real roots if and only if $a>\frac{3}{4}$?

I can show that if the polynomial has four distinct real roots then $a>\frac{3}{4}$ as follows: We first note that if $a<0$ then there are clearly no non-negative roots, and Descartes' rule of signs would show us that there are at most two negative roots, and so we must have that $a > 0$. (Alternatively, we could note that the second derivative must have two distinct roots, which would also tell us that $a>0$)

Now we note that if the polynomial has four distinct real roots, then by Rolle's theorem, its derivative has three distinct real roots. i.e. $$4x^3-4ax+1=0$$ has three distinct real roots. Descartes' rule of signs shows that it has exactly one negative root, and so it must also have a positive root. If $x$ is one of these roots, then we have that $$4ax = 4x^3+1 = 4x^3 + \frac{1}{2} + \frac{1}{2} \geq 3x$$ by AM-GM, and so $a \geq \frac{3}{4}$. For $a=\frac{3}{4}$, we see that the original polynomial has a repeated root at $x=\frac{1}{2}$, and so in fact $a > \frac{3}{4}$.

I have only managed to prove that the converse is true for $a>1$. We can argue as follows: Let $P(x)=x^4-2ax^2+x+a^2-a$. It is then possible to show that

• If $P(x)$ has a repeated root, then this root is equal to $\frac{4a}{4a+3}$, and from here one can show that $a$ must be one of $\{-\frac{1}{4}, 0, \frac{3}{4}\}$. (I have not checked which of these values do in fact give repeated roots, since none of them are larger than $\frac{3}{4}$.) So we need only check that all of the roots of $P$ are real. (The above of course requires some calculations to find)

• If $a>1$ then $P(-\sqrt{a})<0$, $P(0)>0$ and $P(\sqrt{a})<0$, and so $P$ has at least one negative and one positive root. Descartes rule of signs then implies that $P$ has two negative and two positive roots, and so from the above, $P$ has four distinct real roots.

My question is thus the following:

• Can we show that the polynomial has four distinct roots when $a\in(\frac{3}{4}, 1]$ (We should be able to since it is presumably true)

• Is there a way to come to these results without using Calculus? It is after all a high school mathematics contest, and generally problems in such contests meet the requirement that they can be solved without calculus since many of the students would not have done any calculus in school. Moreover, Descartes' rule of signs is not taught in South African high schools at all.

Note that the polynomial factors as $$(x^2+x-a)(x^2-x+1-a)$$ The discriminants are both positive if and only if $a\gt 3/4$. And if $a\gt 3/4$, subtraction shows the two quadratics cannot have a common root. Well within high school range.

• You win. In hindsight, it should have been obvious to try to factor it since we are looking for the roots of the polynomial. – Dylan Sep 12 '15 at 20:36
• How did you factor it? – Ahmed S. Attaalla Sep 13 '15 at 3:10
• It was clear what a factorization had to look like if there was one. For example the $x$ terms had to be the negatives of each other. So I kind of looked at the thing, saw that everything fell into place and wrote it down. – André Nicolas Sep 13 '15 at 3:19

The given equation can be written as $a^2-(2x^2+1)a+x^4+x=0$

$a=\frac{2x^2+1\pm\sqrt{(2x^2+1)^2-4(x^4+x)}}{2}=\frac{2x^2+1\pm(2x-1)}{2}$

$a=x^2+x$ or $a=x^2-x+1$

which are now quadratics in $x$

• This is a cute answer. Changing the focus from a polynomial in $x$ to one in $a$ is very nice. – RKD Jun 19 '17 at 21:20