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We are given four functions:

- \( f(x) = 2x + 6 \)
- \( g(x) = 2x^2 \)
- \( h(x) = 3x \)
- \( k(x) = x + 1 \)

The equation provided is \( f(h(x)) = g(k(x)) \), and we are tasked with finding the values of \( x \) such that this equation holds true.

### Step-by-step solution:

1. **Substitute \( h(x) \) into \( f(x) \):**
   \[
   f(h(x)) = f(3x) = 2(3x) + 6 = 6x + 6
   \]

2. **Substitute \( k(x) \) into \( g(x) \):**
   \[
   g(k(x)) = g(x + 1) = 2(x + 1)^2 = 2(x^2 + 2x + 1) = 2x^2 + 4x + 2
   \]

3. **Set the two expressions equal to each other:**
   \[
   6x + 6 = 2x^2 + 4x + 2
   \]

4. **Rearrange the equation to form a quadratic equation:**
   \[
   2x^2 + 4x + 2 - 6x - 6 = 0
   \]
   Simplifying:
   \[
   2x^2 - 2x - 4 = 0
   \]

5. **Divide the entire equation by 2:**
   \[
   x^2 - x - 2 = 0
   \]

6. **Factor the quadratic equation:**
   \[
   (x - 2)(x + 1) = 0
   \]

7. **Solve for \( x \):**
   \[
   x - 2 = 0 \quad \text{or} \quad x + 1 = 0
   \]
   \[
   x = 2 \quad \text{or} \quad x = -1
   \]

### Conclusion:
The values of \( x \) are \( a = 2 \) and \( b = -1 \).

Would you like more details or explanations on any step?

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Here are five related questions for further practice:
1. What is the composition of \( f(k(x)) \) for the same functions?
2. How would you solve \( g(h(x)) = f(k(x)) \)?
3. What are the roots of \( 3x^2 - 2x + 1 = 0 \)?
4. If \( h(x) = x^2 \), how does the solution change for \( f(h(x)) = g(k(x)) \)?
5. Can the quadratic formula be applied to solve \( x^2 - x - 2 = 0 \)? How?

**Tip:** Always check if a quadratic equation can be factored easily before applying the quadratic formula.

If f(h(x)) = g(k(x)), then the values of x are a and b. The values of a and b, in that order, are ____.

Learn how to solve the equation f(h(x)) = g(k(x)) by using function composition and solving a quadratic equation. This problem involves functions f(x) = 2x + 6, g(x) = 2x^2, h(x) = 3x, and k(x) = x + 1. Suitable for students in Grades 9-12.

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