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:

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

And so we find the discriminant is:

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

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:

\(\displaystyle k\in\left\{9,\frac{11\pm\sqrt{13}}{2}\right\}\)

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

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

\(\displaystyle x\in\left\{-3,-3\pm\sqrt{1+\sqrt{13}}\right\}\)

 

[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)).

 

[CellCoree]

I would approach this problem by first analyzing the given polynomial $P(x)=x^2+6x+k$. We can see that this is a quadratic equation, which means it will have a maximum of two real roots. However, the number of real roots for $P(P(x))$ will depend on the value of $k$.

To find when $P(P(x))$ will have exactly 3 distinct real roots, we can set up the following inequality:

$P(P(x))>0$

Since $P(x)$ can only have two real roots, we can assume that $P(P(x))$ will have three distinct real roots when $P(x)$ is positive. So, we can solve for the values of $k$ that will make $P(x)$ positive for all real $x$.

To do this, we can use the discriminant of the quadratic formula. The discriminant, $b^2-4ac$, will determine the number of real roots a quadratic equation has. If the discriminant is greater than 0, the equation will have two distinct real roots. If the discriminant is equal to 0, the equation will have one real root. And if the discriminant is less than 0, the equation will have no real roots.

In this case, we can set the discriminant of $P(x)$ equal to 0 to find the values of $k$ that will make $P(x)$ positive for all real $x$:

$b^2-4ac=0$

$(6)^2-4(1)(k)=0$

$36-4k=0$

$4k=36$

$k=9$

Therefore, when $k=9$, $P(P(x))$ will have exactly 3 distinct real roots. This is the only value of $k$ that will result in three distinct real roots for $P(P(x))$, as any other value of $k$ will result in either two or no real roots for $P(x)$.

In conclusion, the value of $k$ that will make $P(P(x))$ have exactly 3 distinct real roots is $k=9$. This can be verified by plugging in $k=9$ into $P(x)$ and solving for the roots, which will result in three distinct real roots for $P(P(x))$.