How to find the cubic with roots: $k$, $k^{-1}$ and $1-k$?

This is the second part of a question that asks the same thing but for a quadratic, that part seemed to be fine. The next part asks you to show that: $$x^3-\frac{3}{2}x^2-\frac{3}{2}x+1=0$$ is the only cubic equation of the form $x^3+px^2+qx+r=0$, where $p,q,r \in \mathbb{R}$ which has the following properties: If $k$ is a (possibly complex) root, then $k^{-1}$ is a root and, if $k$ is a root then $1-k$ is a root.

Now the way I intiallly went about it was to write the general cubic as: $$x^3+px^2+qx+r=(x-k)(x-\frac{1}{k})(x-(1-k))$$ And then try to deduce the required equation, though I can't seem to be able to get to it.

A point that confuses me a little is that if we let $k$ be a root then we know that $k^{-1}$ and $1-k$ are also roots. But then couldn't you say that $1-k^{-1}$ must also be a root and consequently so must $1/(1-k^{-1})$. Therefore implying that in order to satisfy the root properties completely there must be 6 roots but obviously this must be wrong.

• How did you decide that $1-k^{-1}$ should be a root? – Ben Grossmann Mar 6 '15 at 19:47
• @Omnomnomnom Since it said that if $k$ is a root then $1-k$ is also a root. Then since we already know that $k^{-1}$ must be a root, then (surely) so must $1-k^{-1}$. But this reasoning must be false but it's not clear to me where. – Jay Mar 6 '15 at 20:04
• You are correct, if $k$ is a root implies $1/k$ and $1-k$ are roots then $k$ is a root implies $1/k$ is a root implies $1-\frac1k$ is a root. – Gregory Grant Mar 6 '15 at 20:05
• A cubic can have at most three roots (a consequence of the Fundamental Theorem of Algebra). If $k$, $1 - k$, and $k^{-1}$ are roots, $1 - k^{-1}$ could only be a root if it were equal to one of the three known roots. – N. F. Taussig Mar 6 '15 at 20:05
• You don't get six roots because $k$ and $\frac1k$ may coincide, for example if $k=-1$. – Gregory Grant Mar 6 '15 at 20:06

If you expand it out you get $$x^3 + px^2 + qx + r = x^3 - \left(1+\frac{1}{k}\right)x^2 + \left(k+\frac{1}{k}-k^2\right)x + k-1$$ Matching terms, you see that $$1+\frac{1}{k} = -p$$ $$k + \frac{1}{k}-k^2 = q$$ $$k-1=r$$ Now, the roots of a cubic with real coefficients must have 3 real roots or 1 real root with a complex conjugate pair. Clearly if $k$ is real, all the roots will be real. If $k$ has nonzero imaginary part, then $1-k$ is also complex and never equal to $\bar{k}$, so $k$ must be real.

As noted in the comments, $1-k^{-1}$ must be a root (and equal to one of the other 3).

If $k < 0$, then $k^{-1} < 0$, $1-k > 0$, and $1-k^{-1} > 0$, suggesting $1-k=1-k^{-1}$, giving $k=1$, a contradiction.

If $0<k<1$, then $0<1-k<1$, $k^{-1}>1$, and $1-k^{-1} < 0$, so $1-k^{-1}$ can't be matched with any root.

We are left with the case $k>1$, so $0<k^{-1}<1$, $1-k < 0$, and $0< 1-k^{-1}<1$, so $1-k^{-1} = k^{-1}$ is the only matching possible, giving $k=2$, which plugging in gives the values of $p,q,r$.

Note that $k=0$ or $k=1$ would result in roots at infinity.

• You can't get a contradiction to the hypothesis $k\lt0$, since $k=-1$ is a root (the other two being $k=1/2$ and $k=2$). – Barry Cipra Mar 7 '15 at 3:31

Hint $\$ Since the roots are closed under reciprocation, the polynomial equals its reciprocal (reverse), thus $\,f = x^3 + px^2+px+1.\,$ Since $\,g(x) = -f(1\!-\!x)\,$ has the same roots and same leading term, it has same constant term $\, 1 = g(0) = -f(1) = -2p-2,\,$ so $\ p=-3/2.$

By comparing coefficients you obtain: $r = k-1,q=1+k(1-k)+\frac{1-k}{k},p=-k-k^{-1}-1+k=k^{-1}-1$. Then $k=r+1$ from which follows $p=-1-(r+1)^{-1}$ and $q=1-r(r+1)- \frac{r}{r+1}$.

Now it must be shown that it must hold $r=1$.

Because $1-k^{-1}$ is also a root (since Substitution of $k \mapsto k^{-1}$ into the root $1-k$ obtains this result); this root must be equal to $k^{-1}$. This is because there are already 2 roots $k$ and $k-1$ given. Now, $1-k^{-1}=k^{-1}$ is satisfied only for $k=2$ whence $r=1$.

Your concern led you to the right track. The roots of the cubic must be $$\{ k, \frac{1}{k}, 1-k, \frac{k-1}{k}, \frac{1}{1-k},\frac{k}{k-1} \}$$ You can verify that applying the requirement to any of the last three or those you just get other members of that same set.

So only very special values of $k$ will work, values that cause those six expressions to evaluate to just three numbers. It turns out that the only solutions that work and also give the product of the roots as a real number are euqivalent to the one generated by $k=\frac{1}{2}$, which leads to roots $$\{ \frac{1}{2}, 2,-\frac{1}{2}, -1, 2, -1 \} = \{ \frac{1}{2}, 2,-1 \} \}$$ which are the roots of the given answer equation.

Any cubic must have at least one real root $k$. If $k$ is real, then $1/k$ and $1-k$ are also real. But $k$, $1/k$, and $1-k$ cannot all be equal: If $k=1-k$, then $k=1/2$, in which case $1/k=2\not=k$. Thus any cubic with the prescribed property must have at least two real roots, and hence all three roots are real.

You cannot have $k=0$ as a root since $1/0$ is not defined. Nor can $k=1$ be a root, since then $1-k=0$ would be a root. Thus any positive roots come in reciprocal pairs, which means there can be at most one such pair. Therefore there is at least one negative root.

But there cannot be two negative roots: if $k_1$ and $k_2$ are negative, then $1-k_1$ and $1-k_2$ are positive. Thus there is exactly one negative root. At this point everything unravels: If $k$ is the unique negative root, then $1/k=k$, which implies $k=-1$, and thus $1-k=2$ and $1/2$ are the other two roots.