what is the proof for the above equation always having exactly one solution greater than zero for all values of a

I cannot see how to prove this because you cannot factorise the polynomial and I am not sure whether looking at its turning points will help

all I can see is that it crosses the y axis at y=-2

  • $\begingroup$ Plugging in $x=0$ won't tell you anything useful, unless $x=0$ happens to be a root. $\endgroup$ – graydad Nov 3 '14 at 22:01
  • $\begingroup$ Are you thinking of $x^3 + ax^2 -x -2=0$ ? $\endgroup$ – Mikael Jensen Nov 3 '14 at 22:07

Note that if $f(x)=x^3+ax^2-x-1$ you have $f(0)=-2$ and $f'(0)=-1$.

Now for $x$ with large absolute value $f'(x)\approx 3x^2$ is positive. So $f'$ changes sign at least once for negative $x$ and at least once for positive $x$ - and since it has at most two changes of sign, it changes precisely once for positive $x$. This must be at a point for which $f(x)$ is negative.

You should be able to conclude from there.


Suppose $f$ has roots $c_1$, $c_2$, and $c_3$. Then we can write:

$$f(x) = (x-c_1)(x-c_2)(x-c_3)$$

Expanding this out, we get:

$$f(x) = x^3 - (c_1+c_2+c_3)x^2 + (c_1c_2+c_1c_3+c_2c_3)x - c_1c_2c_3$$

Notice that the constant term is the signed product of the roots, which is negative in your problem.

Further, the coefficient on $x$ is negative in your problem.

Combining these facts, what can you deduce about the number of positive roots?

  • $\begingroup$ That is of no consequence for my solution. I am looking only at the constant term and the coefficient on $x$. $\endgroup$ – Kaj Hansen Nov 3 '14 at 22:21
  • $\begingroup$ I see what you are saying now - thanks for taking the time to explain. That's the coefficient which was the clue to the way I answered too - interesting to see that the insights are related. NB I kept looking at the coefficient of $x^2$ rather than $x$ as you spotted - I will be paying more attention in future - you have cured me of a lazy habit. $\endgroup$ – Mark Bennet Nov 3 '14 at 22:34
  • $\begingroup$ No problem @MarkBennet. Your solution is pretty slick itself, and it's interesting to see how the different solutions are subtlely related. (Also, I deleted my previous comment so that OP doesn't have the entire answer all spelled out for him.) $\endgroup$ – Kaj Hansen Nov 3 '14 at 22:42

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