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To solve **Question 9b**, which involves using a substitution to solve the quartic equation, we'll break down the steps carefully. Here's the given equation and instructions:

### Step 1: Given equation
We are given the equation from **9a**:

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

You are asked to solve this by using a substitution in the form:

\[
y = x^2 - px
\]

where **p** is a parameter we need to find, and then substitute **y** into the equation to simplify.

---

Let's solve this equation step by step:

1. **Expand and substitute the quadratic terms**:
   - Express \( y = x^2 - px \) and rewrite the equation in terms of **y**.

2. **Solve the equation** after substituting.

Let me calculate it.Let me further analyze the steps to manipulate the equation effectively. The equation will require systematic substitution of \( y = x^2 - px \), expanding, and simplifying to arrive at a solution for the quartic equation.

I will continue solving it.Here is how we proceed:

1. **Substitute** \( y = x^2 - px \) into the quartic equation \( x^4 - 4x^3 - 7x^2 + 22x + 24 = 0 \).
   
   By substituting this, we simplify the equation to a quadratic form in terms of **y**.

2. **Expand** and simplify the terms to solve for **y** using standard algebraic techniques.

Allow me to compute the result for this substitution.To simplify the quartic equation using the substitution \( y = x^2 - px \), we will follow a series of steps involving algebraic manipulation, expand the terms, and solve for \( y \). Given the complexity of the quartic form, we are essentially trying to convert the quartic

Hence 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 by using the quadratic substitution y = x^2 - px. This problem covers key algebraic concepts and techniques, including substitution and solving quadratic equations, suitable for students in Grades 10-12.

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