A cubic function $f$ is defined for $x \in \mathbb{R}$ as $$f : x \mapsto x^3 + (1 - k^2)x + k,$$ where $k$ is a constant - Leaving Cert Mathematics - Question 3 - 2012

We already established that $-k$ is a root; thus, $(x + k)$ is a factor of $f(x)$. To find the other factor, we perform polynomial long division of
$x^3 + (1 - k^2)x + k$ by $(x + k)$.

After performing the long division, we find that
$f(x) = (x + k)(x^2 - kx + (1 - k^2)).$

To find the other two roots, we set the quadratic equation $x^2 - kx + (1 - k^2) = 0$ and use the quadratic formula:

$x = \frac{k \pm \sqrt{k^2 - 4(1 - k^2)}}{2}.$

This simplifies to:
$x = \frac{k \pm \sqrt{5k^2 - 4}}{2}.$

Thus, the other two roots of the equation are:

1. $\frac{k + \sqrt{5k^2 - 4}}{2}$
2. $\frac{k - \sqrt{5k^2 - 4}}{2}$.

For $f$ to have exactly one real root, the discriminant of the quadratic equation must be zero. From the earlier derived form:
$5k^2 - 4 = 0.$
Solving for $k$, we get:
$5k^2 = 4 \implies k^2 = \frac{4}{5} \implies k = \pm \frac{2}{\sqrt{5}}.$

Additionally, for $f$ to have one real root, we require $k^2 - 4 < 0$, leading to the range
$-2 < k < 2.$

Thus, the set of values of $k$ for which $f$ has exactly one real root is $k \in \left(-\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)$.