Cubic equation - Double root

Question

I've been having trouble with this problem:

For which values of $k$ does the equation have a double root? $$x^3-kx^2+k-1=0$$

The correct answer is: $k=-3$, $\frac{3}{2}$, $1$

Thanks!!!

Answer 1

First I want to point out that your question is insufficiently imprecise. It should ask for the real values of $k$ for which the real polynomial given has real/complex double roots.

An alternative method is to factorize the cubic, but note that this does not work when you cannot factorize, whereas the other method by Arnaldo works in all cases.

$x^3 - k x^2 + k x - 1 = (x-1) ( x^2 + x + 1 ) - k (x-1) (x+1) = (x-1) ( x^2 - (k-1) x - (k-1) )$.

For this cubic to have a double root, either the double root must be $1$, or the quadratic factor must have the double root. In the first case $(x-1)$ must be a factor of $( x^2 - (k-1) x - (k-1) )$, or equivalently $1^2 - (k-1) (1) - (k-1) = 0$. In the second case $( x^2 - (k-1) x - (k-1) )$ must be divisible by $(x-c)^2$ for some real $c$, and since they are both quadratics with leading coefficient $1$, it must be that $x^2 - (k-1) x - (k-1)$ and $(x-c)^2$ are the same polynomials, so we must have $k = 2c+1$ and $c^2 + 2c = 0$.

The above shows what $k$ must be for the cubic to have a double root. You still need to check that the possibilities found actually work.

Answer 2

Hint

If $a$ is a double root then:

$$f(a)=a^3-ka^2+k-1=0$$

$$f'(a)=3a^2-2ka=0=a(3a-2k) \rightarrow a=0 \quad \text{or} \quad a=\frac{2k}{3}$$

Ps.: By the rational root theorem, $1$ is a root of $f(x)$: If $1$ is the double root who is $k$? If not, where is the double root?

Can you finish?