Problem source

The quadratic equation $x^{2}+bx+c=0$, where $b$ and $c$ are real, has the properly that if $k$ is a (possibly complex) root, then $k^{-1}$ is a root. Determine carefully the restriction that this property places on $b$ and $c$. If, in addition to this property, the equation has the further property that if $k$ is a root, then $1-k$ is a root, find $b$ and $c$. 
Show that 
\[
x^{3}-\tfrac{3}{2}x^{2}-\tfrac{3}{2}x+1=0
\]
is the only cubic equation of the form $x^{3}+px^{2}+qx+r=0$, where $p,q$ and $r$ are real, which has both the above properties.

Solution source

Suppose $k$ is a root of our quadratic. There are two possibilities, if $k^{-1} = k$ then we must have $k^2 = 1$ so either $\pm 1$ is a root or we must have $(x-k)(x-k^{-1}) = x^2+bx+c$.

In the first case we can have:

\begin{align*}
x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\
x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\
x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\
\end{align*}

In the other cases, $c = 1$ and $b = k^{-1}+k$. Therefore we must have $c = 1$ and $b$ can take any values.

The statement "if $k$ is a root then $1-k$ is a root" implies these two roots are different, so we must have $1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0$ so $b = -1, c = 1$.

Suppose $x^3+px^2+qx+r = 0$ has the first property, then for any root $k$ we must have:

$k^3 + pk^2 + qk + r = 0$ and $1 + pk^{-1} + qk^{-2} + rk^{-3} = 0$ therefore $x^3+px^2+qx+r$ and $rx^3+qx^2+px+1 = 0$ must have identical roots (since $x = \pm$ either wont work here since they imply having the roots $1, 0$ or $-1, 2, \frac12$ which is exactly our equation. Therefore $r = 1, p = q$.

Suppose $x^3 + px^2 + px+1 = 0$ has the property that if $k$ is a root $1-k$ is a root, therefore:

\begin{align*}
0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\
&= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\
&= -k^3+(3+p)k^2+(-3-3p)k+(2+2p)
\end{align*}

Since these roots must be the same as the original roots, we must have $3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32$