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To solve question 9b, we need to use the substitution \( y = x^2 - px \), where \( p \) and \( q \) are constants, to solve the equation:

\[
x^4 - 4x^3 - 7x^2 + 22x + 24 = 0.
\]

### Step-by-step solution:

1. **Rewriting the equation:**
   The equation is given as:
   \[
   x^4 - 4x^3 - 7x^2 + 22x + 24 = 0.
   \]
   Using the substitution \( y = x^2 - px \), we aim to convert this quartic equation into a quadratic form.

2. **Find \( p \) and \( q \):**
   From part (a) of question 9, we have rewritten the quartic equation in the form:
   \[
   x^4 - 4x^3 - 7x^2 = (x^2 - px)^2 - qx^2,
   \]
   where we expand the squared term to identify \( p \) and \( q \). Let's determine \( p \) and \( q \).

Let me solve for the values of \( p \) and \( q \) next.The expanded expression for \( (x^2 - px)^2 - qx^2 \) is:

\[
x^4 - 2px^3 + (p^2 - q)x^2.
\]

Now, we compare this with the original equation:

\[
x^4 - 4x^3 - 7x^2 + 22x + 24 = 0.
\]

By comparing coefficients:

- For \( x^3 \): \( -2p = -4 \Rightarrow p = 2 \).
- For \( x^2 \): \( p^2 - q = -7 \).

Substituting \( p = 2 \) into the second equation:

\[
2^2 - q = -7 \Rightarrow 4 - q = -7 \Rightarrow q = 11.
\]

Thus, \( p = 2 \) and \( q = 11 \).

### Step 3: Substitute into the quadratic form

Now, substitute \( y = x^2 - 2x \) into the equation to solve:

\[
y^2 - 11y + 24 = 0.
\]

Next, I will solve this quadratic equation for \( y \).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can proceed by solving the quadratic equation \( y^2 - 11y + 24 = 0 \) manually. 

You can factor it as:
\[
(y - 3)(y - 8) = 0,
\]
so the solutions for \( y \) are:
\[
y = 3 \quad \text{or} \quad y = 8.
\]

Now, recall that \( y = x^2 - 2x \), so solve the equations:

1. \( x^2 - 2x = 3 \)
2. \( x^2 - 2x = 8 \)

Would you like to solve these two quadratic equations next?

### Here are five related questions:
1. How do you solve a general quadratic equation by factoring?
2. What is the relationship between a quartic and a quadratic equation when using substitutions?
3. Can substitution methods always simplify polynomial equations?
4. How do you handle complex roots in quadratic equations?
5. What happens if the discriminant of a quadratic equation is negative?

**Tip:** When using substitutions, always check if the transformed equation can be solved using familiar techniques like factoring, completing the square, or the quadratic formula.

Use a substitution of the form y = x^2 - px to solve the equation x^4 - 4x^3 - 7x^2 + 22x + 24 = 0.

Learn how to solve the quartic equation x^4 - 4x^3 - 7x^2 + 22x + 24 = 0 using the substitution y = x^2 - px. This solution transforms the quartic equation into a quadratic form, factoring and solving for y. Suitable for students in Grades 10-12.

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