Further Mathematics (9665) | Roots and coefficients of a quadratic equation Content Additional information

FP1.4: Roots and Coefficients of a Quadratic Equation

Welcome to one of the most powerful and satisfying topics in Further Mathematics! This chapter shows you how to connect the numbers that define a quadratic equation (the coefficients) directly to the properties of its solutions (the roots), even without solving the equation itself.

This skill is essential for problem-solving in FP1. We will learn to manipulate complicated expressions involving roots and then use these manipulated results to build entirely new quadratic equations.

1. The Fundamental Link: Sum and Product of Roots

Every quadratic equation of the form \(ax^2 + bx + c = 0\) (where \(a \neq 0\)) has two roots. In Further Maths, we conventionally call these roots \(\alpha\) (alpha) and \(\beta\) (beta).

The Key Formulas

If \(\alpha\) and \(\beta\) are the roots of \(ax^2 + bx + c = 0\), the relationship between the roots and the coefficients is:

• Sum of the roots: \(\alpha + \beta = -\frac{b}{a}\)
• Product of the roots: \(\alpha\beta = \frac{c}{a}\)

Memory Aid: Think of S.A.P. (Sum, a, minus b) and P.C.A. (Product, c, a). The sum always includes the minus sign from the second coefficient (\(b\)).

Key Takeaway: Before you do anything else in a roots and coefficients problem, always identify the values of \(\alpha + \beta\) and \(\alpha\beta\).

2. Manipulating Symmetric Expressions

The core skill in this chapter is taking complicated expressions involving \(\alpha\) and \(\beta\) (like \(\alpha^2 + \beta^2\) or \(\frac{1}{\alpha} + \frac{1}{\beta}\)) and rewriting them exclusively in terms of the fundamental building blocks: \(\alpha + \beta\) and \(\alpha\beta\). These are called symmetric expressions because swapping \(\alpha\) and \(\beta\) doesn't change the value of the expression.

Step-by-Step Derivations

We use algebraic identities to achieve this manipulation.

A. The Sum of Squares: \(\alpha^2 + \beta^2\)

We know the expansion for the sum squared: \((\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2\).
We want \(\alpha^2 + \beta^2\), so we rearrange the identity:

\[ \mathbf{\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta} \]

Analogy: Think of \(\alpha + \beta\) as a package (S) and \(\alpha\beta\) as a package (P). You are building a complex structure using only these two standard packages.

B. Expressions involving Reciprocals (Fractions)

To manipulate expressions like \(\frac{1}{\alpha} + \frac{1}{\beta}\), we first find a common denominator:

\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha\beta} + \frac{\alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta} \]

This is now fully expressed in terms of Sum and Product, making it easy to calculate.

C. The Sum of Cubes: \(\alpha^3 + \beta^3\) (A common and challenging example)

The key identity for cubes is:

\[ (\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 \]

We group the middle terms:

\[ (\alpha + \beta)^3 = (\alpha^3 + \beta^3) + 3\alpha\beta(\alpha + \beta) \]

Rearranging to find \(\alpha^3 + \beta^3\) (as required by the syllabus):

\[ \mathbf{\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)} \]

Key Takeaway: Any symmetric expression involving \(\alpha\) and \(\beta\) can be written purely using \((\alpha + \beta)\) and \((\alpha\beta)\). Always start by manipulating the complex expression first, then substitute the numerical values.

3. Forming a New Quadratic Equation

Often, you are asked to find a new quadratic equation whose roots are related to the roots of an original equation (e.g., if the original roots are \(\alpha, \beta\), the new roots might be \(\alpha^2, \beta^2\) or \(\frac{1}{\alpha}, \frac{1}{\beta}\)).

The Reverse Engineering Process

A general quadratic equation can always be written in terms of its roots, \(\gamma\) and \(\delta\), as:

\[ x^2 - (\gamma + \delta)x + (\gamma\delta) = 0 \]

Therefore, to find the new equation, you just need two things:

1. The New Sum (\(\gamma + \delta\)).
2. The New Product (\(\gamma\delta\)).

Step-by-Step Example: Roots are \(\alpha^2\) and \(\beta^2\)

Suppose the original equation is \(x^2 - 4x + 3 = 0\). (Here, \(\alpha + \beta = 4\) and \(\alpha\beta = 3\)). We want a new equation with roots \(\gamma = \alpha^2\) and \(\delta = \beta^2\).

Step 1: Calculate the New Sum
New Sum \( = \gamma + \delta = \alpha^2 + \beta^2 \)
Using our identity from Section 2:

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (4)^2 - 2(3) = 16 - 6 = 10 \]

Step 2: Calculate the New Product
New Product \( = \gamma\delta = \alpha^2\beta^2 \)
Since \(\alpha^2\beta^2 = (\alpha\beta)^2\):

\[ \alpha^2\beta^2 = (3)^2 = 9 \]

Step 3: Form the New Equation
Use the template \(x^2 - (\text{New Sum})x + (\text{New Product}) = 0\):

\[ \mathbf{x^2 - 10x + 9 = 0} \]

Syllabus Examples to Practice

The syllabus specifically mentions forming equations with roots such as \(\alpha^2, \beta^2\), \(\alpha^3, \beta^3\), \(\frac{1}{\alpha}, \frac{1}{\beta}\), and composite fraction forms like \(\frac{2}{\beta} + \frac{2}{\alpha}\). Remember that all these roots must first be simplified into combinations of \((\alpha + \beta)\) and \((\alpha\beta)\).

Example: Roots \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)

• New Sum: \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\) (Ratio of original sum to original product)
• New Product: \(\frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta}\) (Reciprocal of original product)

Don't worry if this seems tricky at first; with practice, these manipulations become second nature!

4. Summary of Key Identities

Keep these fundamental algebraic identities handy, as they are your tools for FP1 roots problems:

Let \(S = \alpha + \beta\) and \(P = \alpha\beta\):

• \(\alpha^2 + \beta^2 = S^2 - 2P\)
• \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{S}{P}\)
• \(\alpha^3 + \beta^3 = S^3 - 3PS\)
• \(\alpha - \beta\) (or \(\beta - \alpha\)) is often found using the identity: \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = S^2 - 4P\)

Did you know? The properties you are studying here are the simplest case of Vieta's Formulas, which provide relationships between roots and coefficients for polynomials of any degree (cubic, quartic, etc.). In FP1, we stick strictly to quadratics!