Further Mathematics | Roots of quadratic equations

📚 FP1 Study Notes: Roots of Quadratic Equations

Welcome to one of the most fundamental chapters in Further Pure Mathematics! This topic is elegant because it allows us to know a huge amount about the solutions (roots) of a quadratic equation without ever having to solve it. This saves time and is essential preparation for handling higher-degree polynomials later on. Take a deep breath—we’ll break down these powerful relationships step-by-step!

What is a Root? (Quick Review)

A quadratic equation is typically written as:
\[ ax^2 + bx + c = 0 \] The roots (or solutions) are the values of \(x\) that make this equation true. In FP1, we usually denote these two roots using the Greek letters alpha (\(\alpha\)) and beta (\(\beta\)).

Did you know? The study of relationships between roots and coefficients is the foundation of Galois Theory, one of the most beautiful and complex areas of abstract algebra!

Section 1: The Core Relationships

The key to this chapter is relating the sum and product of the roots directly to the coefficients (\(a\), \(b\), and \(c\)) of the quadratic equation.

1.1 Standard Form and Key Coefficients

To derive the formulas, we first divide the entire equation by \(a\) (assuming \(a \ne 0\)) so that the coefficient of \(x^2\) is 1: \[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

We also know that if \(\alpha\) and \(\beta\) are the roots, the equation can be written in factor form: \[ (x - \alpha)(x - \beta) = 0 \] Expanding this gives: \[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]

1.2 The Formulas (Compare Coefficients)

By comparing the two boxed forms above, we establish the two essential relationships you must memorise:

Accessibility Tip: These formulas allow you to calculate \(\alpha + \beta\) and \(\alpha\beta\) immediately upon seeing an equation. If the equation is \(3x^2 - 6x + 5 = 0\), then \(a=3\), \(b=-6\), \(c=5\).
Sum: \(\alpha + \beta = -(-6)/3 = 6/3 = 2\).
Product: \(\alpha\beta = 5/3\).

Section 2: Manipulating Expressions Involving Roots

A standard FP1 question involves finding the value of complex expressions like \(\alpha^2 + \beta^2\) or \(\frac{1}{\alpha} + \frac{1}{\beta}\), using only the sum (\(\alpha + \beta\)) and the product (\(\alpha\beta\)).

2.1 Essential Derived Identity: \(\alpha^2 + \beta^2\)

We rely on algebraic identities learned in previous courses. The most important one is based on squaring the sum of the roots: \[ (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \] To find \(\alpha^2 + \beta^2\), we rearrange this identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Why is this useful? Because the expression on the right-hand side is only made up of the Sum and the Product, both of which we know directly from the coefficients!

2.2 Handling Other Expressions

To find expressions involving fractions or higher powers, always follow the rule: Rewrite the expression in terms of \(\alpha + \beta\) and \(\alpha\beta\).

Example 1: The Sum of Reciprocals

Find \(\frac{1}{\alpha} + \frac{1}{\beta}\).

Step 1: Combine the fractions using a common denominator (\(\alpha\beta\)): \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha\beta} + \frac{\alpha}{\alpha\beta} \] Step 2: Simplify the numerator: \[ \frac{\beta + \alpha}{\alpha\beta} \] This is now simply \(\frac{\text{Sum}}{\text{Product}}\)!

Example 2: Higher Powers

Find \(\alpha^3 + \beta^3\). (This requires slightly more advanced factoring, but follows the same principle.)
We use the factoring identity: \[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \] Step 1: Substitute the identity for \(\alpha^2 + \beta^2\): \[ \alpha^3 + \beta^3 = (\alpha + \beta) \left[ (\alpha^2 + \beta^2) - \alpha\beta \right] \] \[ \alpha^3 + \beta^3 = (\alpha + \beta) \left[ (\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta \right] \] Step 2: Simplify: \[ \alpha^3 + \beta^3 = (\alpha + \beta) \left[ (\alpha + \beta)^2 - 3\alpha\beta \right] \] Since both components are known (Sum and Product), we can now substitute values.

Section 3: Forming New Equations from Transformed Roots

The culmination of this topic is finding a new quadratic equation whose roots are related to the original roots \(\alpha\) and \(\beta\).

3.1 The Rule for Forming a Quadratic

Any quadratic equation \(x^2 + Bx + C = 0\) can be formed if you know its roots, \(\gamma\) and \(\delta\).
The rule is: \[ x^2 - (\text{Sum of New Roots})x + (\text{Product of New Roots}) = 0 \]

This is the key structure. Our entire task is therefore to calculate the New Sum and the New Product using the known original sum \(S\) and product \(P\).

3.2 Step-by-Step Process for Transformation

Suppose the original equation is \(x^2 - 5x + 3 = 0\). (Original \(S=5\), Original \(P=3\)). We want a new equation whose roots are \(\gamma = 2\alpha\) and \(\delta = 2\beta\).

1. Determine Original Sum (S) and Product (P)

   \(\alpha + \beta = S = 5\)
   \(\alpha\beta = P = 3\)

2. Calculate the New Sum (\(S'\))

   The new roots are \(2\alpha\) and \(2\beta\).
   \(S' = 2\alpha + 2\beta\)
   \(S' = 2(\alpha + \beta) = 2(S)\)
   \(S' = 2(5) = 10\)

3. Calculate the New Product (\(P'\))

   The new roots are \(2\alpha\) and \(2\beta\).
   \(P' = (2\alpha)(2\beta)\)
   \(P' = 4\alpha\beta = 4(P)\)
   \(P' = 4(3) = 12\)

4. Form the New Equation

   Use the structure \(x^2 - S'x + P' = 0\): \[ x^2 - (10)x + (12) = 0 \]

3.3 Handling Complex Transformations (E.g., Squared Roots)

Suppose the new roots are \(\gamma = \alpha^2\) and \(\delta = \beta^2\).

1. New Sum (\(S'\)):

   \(S' = \alpha^2 + \beta^2\)
   We use the identity from Section 2:
   \(S' = (\alpha + \beta)^2 - 2\alpha\beta = S^2 - 2P\)

2. New Product (\(P'\)):

   \(P' = (\alpha^2)(\beta^2)\)
   We can rewrite this using index laws:
   \(P' = (\alpha\beta)^2 = P^2\)

3. Form the Equation:

   \[ x^2 - (S^2 - 2P)x + P^2 = 0 \] (You would substitute the numerical values of S and P from the original equation here.)

Section 4: Alternative Method – Substitution (For Able Students)

For certain transformations, there is an alternative method that avoids calculating the new sum and product explicitly. This method works best when the transformation involves simple mapping, like \(x \to x+k\) or \(x \to kx\).

4.1 The Substitution Principle

If the roots of the new equation are related to the old roots (\(\alpha\), \(\beta\)) by a relationship \(y = f(\alpha)\), we can find the new equation by substituting the inverse function of \(f\) back into the original equation.

Example: Roots are \(\alpha - 3\) and \(\beta - 3\)

The new roots are \(y = x - 3\). (Where \(x\) is the original root \(\alpha\) or \(\beta\)).
The inverse relationship is \(x = y + 3\).

If the original equation is \(ax^2 + bx + c = 0\), we substitute \(x = (y+3)\) into it: \[ a(y+3)^2 + b(y+3) + c = 0 \] Expanding this expression will give the new quadratic equation (in terms of \(y\), which you can then replace with \(x\)).

Encouragement: Don't worry if this substitution method seems tricky at first. It’s an efficient shortcut, but the primary method (finding New Sum and New Product) will always work and is generally safer for complex transformations.

Section 5: Summary and Final Tips

5.1 Checklist for Success

• Check Coefficients: When calculating \(S = -b/a\) and \(P = c/a\), always ensure the original equation is set to zero and correctly identify \(a\), \(b\), and \(c\).
• Mind the Minus: Double-check the sign of the Sum (\(-b/a\)). It is the most common place to drop marks.
• Use the Identities: Never try to calculate \(\alpha\) and \(\beta\) themselves! Always use the derived identities like \(\alpha^2 + \beta^2 = S^2 - 2P\).
• New Equation Structure: Always write your final answer in the form \(x^2 - (\text{New Sum})x + (\text{New Product}) = 0\). (Note the minus sign before the sum!)

5.2 Sigma Notation (\(\sum\))

In FP1, you may see expressions written using sigma notation. This is simply a shorthand for the sum of all possible terms created by the expression for each root.

• \(\sum \alpha = \alpha + \beta\) (The Sum)
• \(\sum \alpha^2 = \alpha^2 + \beta^2\)
• \(\sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta}\)

The techniques used to evaluate these expressions remain exactly the same as in Section 2.

Keep practicing those algebraic manipulations, and you will find that the Roots of Quadratics chapter is a highly rewarding part of FP1! Good luck!