Solving for $k$: When Does $P(P(x))$ Have 3 Real Roots?

[anemone]

Let $P(x)=x^2+6x+k$ for all real $x$, where $k$ is some real number. For what values of $k$ does $P(P(x))$ have exactly 3 distinct real roots?

 

[RLBrown]

Spoiler

WLOG: Let x=y-3

G(y)=P(P(y))= y^4+(2k-12)y^2+h(k)
where h(k) = k^2-11k+21

Note 1: G(y) symmetrical about y=0
Note 2: G'(0)=0

Therefore: Answer is root(s) of h(k)

 

[MarkFL]

My solution:

Spoiler

We find:

$$P(P(x))=\left(x^2+6x+k\right)^2+6\left(x^2+6x+k\right)+k=x^4+12x^3+2kx^2+42x^2+12kx+36x+k^2+7k$$

And so we find the discriminant is:

$$256k^4-7424k^3+78336k^2-352512k+559872=256(k-9)^2\left(k^2-11 k+27\right)$$

In order for there to be 3 distinct real roots, we must have 4 real roots, two of which are repeated, and so the discriminat must be zero.

Thus, we have the candidates:

$$k\in\left\{9,\frac{11\pm\sqrt{13}}{2}\right\}$$

When $k=9$, we find a repeated set of complex conjugate roots. When $$k=\frac{11+\sqrt{13}}{2}$$, we find that we have a repeated real root and a pair of complex conjugate roots.

But, when $$k=\frac{11-\sqrt{13}}{2}$$, we find we have the 3 distinct real roots:

$$x\in\left\{-3,-3\pm\sqrt{1+\sqrt{13}}\right\}$$

 

[RLBrown]

Spoiler

WLOG: Let x=y-3

G(y)=P(P(y))= y^4+(2k-12)y^2+h(k)
where h(k) = k^2-11k+21

Note 1: G(y) symmetrical about y=0
Note 2: G'(0)=0

Therefore: Answer is root(s) of h(k)

G(y)=P(P(y))= y^4+(2k-12)y^2+h(k)
where h(k) = k^2-11k+27 <--- Typo correction

Note 1: G(y) symmetrical about y=0
Note 2: G'(0)=0

Therefore: Answer is root(s) of h(k)

As MarkFL points out, only the smaller root of h(k) produces 3 roots in G(y) and P(P(x)).
The larger root of h(k) produces only one double root in G(y) and P(P(x)).