Given that the equation x² + 2x + 4 = P(x² + 4) has equal roots, find the possible values of P.

The given equation is:

$$x^2 + 2x + 4 = P(x^2 + 4)$$

Distribute $P$ on the right side:

$$x^2 + 2x + 4 = Px^2 + 4P$$

Move all terms to one side to form a standard quadratic equation:

$$x^2 - Px^2 + 2x + 4 - 4P = 0$$

Factor out $x^2$:

$$(1-P)x^2 + 2x + (4-4P) = 0$$

From the standard quadratic form $ax^2 + bx + c = 0$, we have:

$$a = 1-P$$ $$b = 2$$ $$c = 4-4P$$

For a quadratic equation to have equal roots, its discriminant must be zero. The discriminant is given by $b^2 - 4ac$.

$$b^2 - 4ac = 0$$

$$(2)^2 - 4(1-P)(4-4P) = 0$$ $$4 - 4(1-P) \cdot 4(1-P) = 0$$ $$4 - 16(1-P)^2 = 0$$

Divide the entire equation by 4:

$$1 - 4(1-P)^2 = 0$$ $$1 = 4(1-P)^2$$ $$\frac{1}{4} = (1-P)^2$$

Take the square root of both sides:

$$\pm\sqrt{\frac{1}{4}} = 1-P$$ $$\pm\frac{1}{2} = 1-P$$

Case 1:

$$\frac{1}{2} = 1-P$$ $$P = 1 - \frac{1}{2}$$ $$P = \frac{1}{2}$$

Case 2:

$$-\frac{1}{2} = 1-P$$ $$P = 1 + \frac{1}{2}$$ $$P = \frac{3}{2}$$

The possible values of $P$ are $\frac{1}{2}$ and $\frac{3}{2}$. Note that for these values, $a = 1-P \neq 0$, so the equation remains quadratic.

The possible values of $P$ are $\boxed{\frac{1}{2}, \frac{3}{2}}$.