Question: If one root of the equation $x^2 + px + 12 = 0$ is $4$, while the equation $x^2 + px + q = 0$ has equal roots, then $q =$
○ $3$
○ $12$
○ $\frac{49}{4}$
○ $4$
1. **State the problem:**
We have two quadratic equations:
$$x^2 + px + 12 = 0$$
and
$$x^2 + px + q = 0$$
Given that one root of the first equation is $4$, and the second equation has equal roots, we need to find the value of $q$.
2. **Use the root to find $p$:**
Since $4$ is a root of $x^2 + px + 12 = 0$, substitute $x=4$:
$$4^2 + p(4) + 12 = 0$$
$$16 + 4p + 12 = 0$$
$$4p + 28 = 0$$
$$4p = -28$$
$$p = -7$$
3. **Use the condition for equal roots in the second equation:**
For the quadratic equation $x^2 + px + q = 0$ to have equal roots, its discriminant must be zero:
$$\Delta = p^2 - 4q = 0$$
Substitute $p = -7$:
$$(-7)^2 - 4q = 0$$
$$49 - 4q = 0$$
$$4q = 49$$
$$q = \frac{49}{4}$$
4. **Final answer:**
$$q = \frac{49}{4}$$