TMUA practice: Polynomials, problem 5

The depressed cubic $\;x^{3} + p x + q = 0\;$ has a repeated real root iff its discriminant $\;-4 p^{3} - 27 q^{2}\;$ vanishes, i.e.

$4 p^{3} + 27 q^{2} \;=\; 0.$

Here $p = -1$, $q = a$, so $\;-4 + 27 a^{2} = 0$, giving $\;a^{2} = 4/27$ and $\;a = \pm \dfrac{2}{3\sqrt{3}} = \pm \dfrac{2\sqrt{3}}{9}$.

Two distinct real values. This is option C.

Alternative: $f(x) = x^{3} - x + a$ has $f'(x) = 3 x^{2} - 1 = 0$ at $x = \pm 1/\sqrt{3}$. Repeated root occurs when $f$ has a double root, which happens at the local extrema. $f(1/\sqrt{3}) = 0$ gives $a = 2/(3\sqrt{3})$; $f(-1/\sqrt{3}) = 0$ gives $a = -2/(3\sqrt{3})$. Two values.

Why the other options fail. Discriminant analysis gives exactly $2$ solutions.

The lesson: a depressed cubic $x^{3} + p x + q$ has a repeated real root iff the discriminant vanishes; in the symmetric case (only $p, q$ nonzero), this gives $\pm$ pairs.