Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $\alpha\in(0,\frac12)$, $\beta\in(0,\infty)$, $N\in\mathbb N\backslash\{0\}$ and $n\in\{0,\ldots,N\}$, how can I prove that exactly one zero of the cubic polynomial
$$ (N+2\beta)x^3-(N+n+3\beta)x^2+(n+\beta+N\alpha-N\alpha^2)x+n\alpha^2-n\alpha $$ lies in $[\alpha,1-\alpha]$?

share|cite|improve this question
I would evaluate the polynomial at $\alpha$ and $1-\alpha$, hoping to find different signs, and then I would compute the discriminant, hoping to find zero. Did you try? – Giovanni De Gaetano Feb 9 '12 at 17:30
up vote 4 down vote accepted

Evaluating at $\alpha$, we have: $$\begin{align*} f(\alpha) &= (N+2\beta)\alpha^3 - (N+n+3\beta)\alpha^2 + (n+\beta+N\alpha-N\alpha^2)\alpha + n\alpha^2 - n\alpha\\ &= (N+2\beta - N)\alpha^3 + (-N-n-3\beta+N+n)\alpha^2 + (n+\beta-n)\alpha\\ &= 2\beta\alpha^3 - 3\beta\alpha^2 + \beta\alpha\\ &= \alpha\beta(2\alpha^2 -3\alpha + 1). \end{align*}$$ Evaluating at $1-\alpha$ gives $$\begin{align*} f(1-\alpha) &= (N+2\beta)(1-3\alpha + 3\alpha^2-\alpha^3) - (N+n+3\beta)(1-2\alpha+\alpha^2)\\ &\qquad \mathop{+} (n+\beta+N\alpha-N\alpha^2)(1-\alpha) + n\alpha^2 - n\alpha\\ &= (-N-2\beta +N)\alpha^3 + (3N+6\beta - N-n-3\beta -N-N+n)\alpha^2\\ &\qquad \mathop{+}(-3N-6\beta+2N+2n+6\beta-n-\beta+N-n)\alpha\\ &\qquad \mathop{+} (N+2\beta-N-n-3\beta+n+\beta)\\ &= -2\beta\alpha^3 +3\beta\alpha^2-\beta\alpha\\ &= -\alpha\beta(2\alpha^2 - 3\alpha + 1). \end{align*}$$ So, unless $2\alpha^2-3\alpha+1$ is $0$, the two values have opposite signs. But the roots of $2x^2-3x+1$ are $1$ and $\frac{1}{2}$, so $\alpha$ cannot be a root.

Thus, there is at least one root for the polynomial in $[\alpha,1-\alpha]$ (in fact, in $(\alpha,1-\alpha)$.

Since $f(x)$ has opposite signs on $\alpha$ and on $1-\alpha$, if $f(x)$ has more than one (distinct) root on $[\alpha,1-\alpha]$, then it must have three distinct roots in the interval (why?). Can all three roots be in that interval?

share|cite|improve this answer
Great! The discriminant is identically zero so that implies that there is at most two distinct roots. But if there were two distinct roots in $(\alpha,1-\alpha)$, it could not change sign. So there is exactly one root in the interval. – Chris Ferrie Feb 9 '12 at 19:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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