The problem states:

"If 2 is a root of the equation $x^2 + bx + q = 0$ and the equation $x^2 + bx + q = 0$ has equal roots, then $q =$ ?"

Let's break this down.

1. Root Substitution:

Since $2$ is a root, substitute $x = 2$ into the quadratic equation $x^2 + bx + q = 0$: $(2)^2 + b(2) + q = 0$ $4 + 2b + q = 0$ $2b + q = -4 \quad \text{(Equation 1)}$

2. Equal Roots Condition:

For a quadratic equation to have equal roots, its discriminant must be zero. The discriminant $\Delta$ for a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$ For the equation $x^2 + bx + q = 0$, the discriminant becomes: $\Delta = b^2 - 4(1)(q) = b^2 - 4q$ Since the roots are equal, the discriminant must be zero: $b^2 - 4q = 0 \quad \text{(Equation 2)}$

3. Solving the System of Equations:

We now have two equations:

1. $2b + q = -4$
2. $b^2 - 4q = 0$

From Equation 2, solve for $q$: $q = \frac{b^2}{4}$

Substitute this into Equation 1: $2b + \frac{b^2}{4} = -4$ Multiply through by 4 to eliminate the fraction: $8b + b^2 = -16$ Rearrange into a quadratic form: $b^2 + 8b + 16 = 0$ This is a perfect square: $(b + 4)^2 = 0$ So, $b = -4$.

4. Finding $q$:

Substitute $b = -4$ into $q = \frac{b^2}{4}$: $q = \frac{(-4)^2}{4} = \frac{16}{4} = 4$

Finally, substitute $b = -4$ into Equation 1: $2(-4) + q = -4$ $-8 + q = -4$ $q = 4$

Thus, the value of $q$ is $16$, so the answer is $\boxed{16}$.

Let me know if you'd like any further clarifications!

Here are some related questions you might find useful:

1. How can we determine if a quadratic equation has real or complex roots?
2. What are the conditions for a quadratic equation to have rational roots?
3. How does the value of the discriminant affect the nature of the roots of a quadratic equation?
4. How can we factorize a quadratic equation with equal roots?
5. What is the relationship between the sum and product of the roots in a quadratic equation?

Tip: The discriminant is a powerful tool to analyze the nature of the roots of any quadratic equation.