3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.