# Functions of the roots of a quadratic equation

> Edexcel International GCSE Further Pure Mathematics · 4PM1 S2
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s2-functions-of-the-roots-of/

This guide covers core skills for working with quadratic roots for Edexcel IGCSE Further Pure Math 4PM1: using sum and product of roots identities, evaluating symmetric functions, and forming new quadratics with transformed roots.

**Prerequisites:** [Solving quadratic equations and discriminant](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s2-quadratic-discriminant-solving/)

## Learning objectives

- Recall and apply the sum ($\alpha + \beta = -b/a$) and product ($\alpha\beta = c/a$) identities for roots of $ax^2 + bx + c = 0$
- Evaluate common symmetric functions of roots including $\alpha^2 + \beta^2$, $1/\alpha + 1/\beta$, $\alpha^3 + \beta^3$ and $(\alpha - \beta)^2$
- Form new quadratic equations with transformed roots using the standard $x^2 - (sum)x + product = 0$ form

## Sum and Product of Quadratic Roots Identities

For any quadratic equation of the form $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, two core identities link the coefficients of the quadratic to its roots. These formulas are not provided in your exam formula sheet, so you must memorize them.

**Sum and Product of Quadratic Roots** — If $ax^2 + bx + c = 0$ has roots $\alpha, \beta$: <br> Sum of roots: $\alpha + \beta = -\frac{b}{a}$ <br> Product of roots: $\alpha\beta = \frac{c}{a}$

*Example:* For $2x^2 + 5x - 3 = 0$, $\alpha + \beta = -5/2$, $\alpha\beta = -3/2$

**Worked example:** Find the sum and product of the roots of $3x^2 -7x + 2 = 0$.

1. Step 1: Identify coefficients: $a=3$, $b=-7$, $c=2$
2. Step 2: Apply sum of roots formula:
3. $$\alpha + \beta = -\frac{b}{a} = -\left(\frac{-7}{3}\right) = \frac{7}{3}$$
4. Step 3: Apply product of roots formula:
5. $$\alpha\beta = \frac{c}{a} = \frac{2}{3}$$

> **warning**
>
> Always check the sign of $b$ when calculating the sum of roots: forgetting the leading negative sign in $\alpha + \beta = -b/a$ is the most common error on this topic.

*Calculator:* allowed

## Evaluating Symmetric Functions of Roots

Symmetric functions of roots are expressions where swapping $\alpha$ and $\beta$ leaves the value unchanged. You do not need to calculate individual roots to evaluate these: all can be rewritten using only the sum and product of roots identities.

- $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
- $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$
- $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$
- $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$

**Worked example:** Given $x^2 -4x +1 = 0$ has roots $\alpha, \beta$, find the value of $\alpha^3 + \beta^3$.

1. Step 1: Calculate sum and product of original roots:
2. $$\alpha + \beta = 4, \quad \alpha\beta = 1$$
3. Step 2: Substitute into the sum of cubes identity:
4. $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$
5. $$= 4^3 - 3(1)(4) = 64 - 12 = 52$$

> **Exam tip**
>
> Never waste time solving for individual $\alpha$ and $\beta$ values for these questions: using the symmetric identities is faster and far less error-prone.

*Calculator:* allowed

## Forming New Quadratics with Transformed Roots

Once you can calculate the sum and product of a transformed set of roots, you can write the full quadratic equation for those roots using a standard form.

**Quadratic Equation from Roots** — If a quadratic has roots $p$ and $q$, its standard form is: <br> $x^2 - (p + q)x + pq = 0$ <br> Multiply through by a constant to clear fractions if integer coefficients are required.

*Example:* A quadratic with roots 3 and 5 has equation $x^2 - 8x +15 = 0$

**Worked example:** The quadratic $2x^2 +3x -2 =0$ has roots $\alpha, \beta$. Form a quadratic with integer coefficients that has roots $2\alpha +1$ and $2\beta +1$.

1. Step 1: Calculate original sum and product of roots:
2. $$\alpha + \beta = -\frac{3}{2}, \quad \alpha\beta = -1$$
3. Step 2: Calculate sum of new roots $S$:
4. $$S = (2\alpha +1) + (2\beta +1) = 2(\alpha + \beta) + 2 = 2\left(-\frac{3}{2}\right) + 2 = -1$$
5. Step 3: Calculate product of new roots $P$:
6. $$P = (2\alpha +1)(2\beta +1) = 4\alpha\beta + 2(\alpha + \beta) +1 = 4(-1) + 2\left(-\frac{3}{2}\right) +1 = -6$$
7. Step 4: Substitute into standard quadratic form:
8. $$x^2 - Sx + P = x^2 +x -6 = 0$$

> **tip**
>
> Always check the question for integer coefficient requirements: multiply all terms by the lowest common denominator of any fractions to get valid integer coefficients.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using $\alpha + \beta = b/a$ instead of $-b/a$
  - Why it fails: Forgetting the leading negative sign in the sum of roots identity invalidates all subsequent calculations
  - Correct: Explicitly memorize the identity with the negative sign, and double-check the sign of $b$ when substituting values
- **Wrong:** Calculating individual $\alpha$ and $\beta$ values to evaluate symmetric functions
  - Why it fails: This is time-consuming and increases risk of arithmetic errors, especially with irrational roots
  - Correct: Always rewrite symmetric functions using only the sum and product of roots identities
- **Wrong:** Writing new quadratics as $x^2 + (sum)x + product = 0$
  - Why it fails: Mixing up the sign of the $x$ term when forming a quadratic from roots leads to an incorrect final equation
  - Correct: Recall the standard form: subtract the sum of roots, add the product of roots
- **Wrong:** Leaving fractions in the final quadratic when integer coefficients are required
  - Why it fails: Exam questions often explicitly require integer coefficients, so un-cleared fractions lose marks even if sum/product are correct
  - Correct: Multiply all terms by the lowest common denominator of any fractions to get integer coefficients
- **Wrong:** Expanding $(\alpha - \beta)^2$ as $\alpha^2 - \beta^2$
  - Why it fails: Confusing the square of a difference with the difference of squares leads to an incorrect symmetric identity
  - Correct: Use the pre-derived identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ to avoid expansion errors

## Cheatsheet

| Identity Name | Formula |
| --- | --- |
| Sum of roots | $\alpha + \beta = -\frac{b}{a}$ |
| Product of roots | $\alpha\beta = \frac{c}{a}$ |
| Sum of squares | $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ |
| Sum of reciprocals | $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$ |
| Sum of cubes | $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ |
| Square of difference | $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ |
| Quadratic from roots | $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ |

## What's next

Now that you have mastered functions of quadratic roots, you are ready to move on to more advanced quadratic function topics in Edexcel IGCSE Further Pure Math 4PM1. This skill is foundational for many algebra topics you will encounter if you progress to A Level Further Mathematics, but for your 4PM1 exam, you should practice applying these identities to a range of exam-style questions, including multi-step problems that combine this topic with other algebra skills. Make sure you memorize all the core identities covered, as they are not provided on your formula sheet, and practice forming quadratics with different transformed root types to build confidence for your exam.

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