To solve the problem, let’s break it down step by step:

We are given:

• $f(x) = x^2 - 3x + 2$
• $g(x) = 2x + 1$
• $(f \circ g)(a) = 12$, meaning $f(g(a)) = 12$.

Step 1: Calculate $g(a)$

The function $g(x) = 2x + 1$, so: $g(a) = 2a + 1$

Step 2: Substitute $g(a)$ into $f(x)$

Substitute $g(a) = 2a + 1$ into $f(x)$: $f(g(a)) = f(2a + 1) = (2a + 1)^2 - 3(2a + 1) + 2$

Expand $(2a + 1)^2$: $(2a + 1)^2 = 4a^2 + 4a + 1$

Substitute into $f(g(a))$: $f(2a + 1) = (4a^2 + 4a + 1) - 3(2a + 1) + 2$

Simplify: $f(2a + 1) = 4a^2 + 4a + 1 - 6a - 3 + 2$ $f(2a + 1) = 4a^2 - 2a + 0$ $f(2a + 1) = 4a^2 - 2a$

Step 3: Set $f(g(a)) = 12$

We know $f(g(a)) = 12$, so: $4a^2 - 2a = 12$

Divide through by 2: $2a^2 - a = 6$

Rearrange into standard quadratic form: $2a^2 - a - 6 = 0$

Step 4: Solve the quadratic equation

The quadratic equation is: $2a^2 - a - 6 = 0$

Use the quadratic formula: $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 2$
• $b = -1$
• $c = -6$

Substitute: $a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-6)}}{2(2)}$ $a = \frac{1 \pm \sqrt{1 + 48}}{4}$ $a = \frac{1 \pm \sqrt{49}}{4}$ $a = \frac{1 \pm 7}{4}$

Solve for the two possible values of $a$: $a = \frac{1 + 7}{4} = \frac{8}{4} = 2$ $a = \frac{1 - 7}{4} = \frac{-6}{4} = -\frac{3}{2}$

Step 5: Verify the solution

The problem asks for $a$ such that $f(g(a)) = 12$. Since $a = 2$ is among the choices, the correct answer is: $\boxed{2}$

Do you want further clarification or more examples?
Here are related questions you might consider:

1. How do you compose two functions step by step?
2. What are the key steps in solving quadratic equations?
3. How does function composition apply to real-world problems?
4. Can we use graphing to verify the solutions for $f(g(a)) = 12$?
5. How would the solution change if the functions $f(x)$ or $g(x)$ were different?

Tip: Always verify your solutions by substituting them back into the original equation to ensure accuracy!