# Quadratic Equations

> Edexcel International GCSE Mathematics A · 4MA1
> Source: https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-quadratic-equations/

This guide covers all quadratic equation content for Edexcel IGCSE Mathematics A (4MA1) specification point 2.7, including Foundation and Higher tier methods, contextual problem solving, and linear-quadratic simultaneous systems.

**Prerequisites:** [Expanding and simplifying algebraic brackets](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-expanding-brackets/); [Factorising linear algebraic expressions](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-factorising-linear-expressions/); [Solving linear simultaneous equations](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-linear-simultaneous-equations/)

## Learning objectives

- Solve Foundation tier quadratic equations of the form $x^2 + bx + c = 0$ by factorisation
- Solve Higher tier general quadratic equations by factorisation, quadratic formula, or completing the square
- Form and solve quadratic equations from real-world contextual data
- Solve linear-quadratic simultaneous equation systems with two unknowns
- Present exact solutions (surds/fractions) unless decimal accuracy is explicitly requested

## Solving Foundation Tier Quadratic Equations ($x^2 + bx + c = 0$)

**Quadratic equation (Foundation form)** — A quadratic equation with leading coefficient 1, written as $x^2 + bx + c = 0$, solvable by factorisation for Foundation tier.

For Foundation tier, you only need to solve quadratics where the coefficient of $x^2$ is 1. First rearrange all terms to the left-hand side so the right-hand side equals 0. Find two integers that multiply to $c$ and add to $b$, then rewrite the quadratic as a product of two linear brackets. Set each bracket equal to 0 to find the two roots.

**Worked example:** Solve $x^2 + x - 30 = 0$

1. Confirm the equation is in standard form with RHS = 0: $a=1$, $b=1$, $c=-30$
2. Find two integers that multiply to -30 and add to 1: 6 and -5
3. Factor the left-hand side: $(x + 6)(x - 5) = 0$
4. Set each bracket equal to 0 and solve: $x + 6 = 0 \implies x = -6$, $x - 5 = 0 \implies x = 5$

> **Exam tip:** Always double check the sign of your factor pairs: if $c$ is negative, your factors will have opposite signs, if $c$ is positive they will have the same sign as $b$.

## Higher Tier General Quadratic Factorisation ($a \neq 1$)

For Higher tier, you may be asked to factorise quadratics where the leading coefficient $a$ is not 1, or quadratics that are given in expanded non-standard form (e.g. with brackets). First expand any brackets and rearrange all terms to get a standard form equation equal to 0. You can use inspection or grouping to factor the expression.

**Worked example:** Solve $x(3x - 2) = 5$

1. Expand and rearrange to standard form: $3x^2 - 2x - 5 = 0$
2. Find two integers that multiply to $a \times c = 3(-5) = -15$ and add to $b=-2$: -5 and 3
3. Split the middle term and factor by grouping: $3x^2 + 3x -5x -5 = 3x(x+1) -5(x+1) = (3x -5)(x+1) = 0$
4. Solve for $x$: $3x -5 = 0 \implies x = \frac{5}{3}$, $x + 1 = 0 \implies x = -1$

> **Exam tip:** If you struggle to factor a quadratic quickly, you can always use the quadratic formula instead -- no factorisation is required to use this method.

## Solving Quadratics Using Formula or Completing the Square

**Quadratic Formula** — A formula that calculates the roots of any standard form quadratic equation, provided on the Higher tier formula sheet.

*Notation:* x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

When factorisation is not possible or too slow, use the quadratic formula or completing the square. The quadratic formula works for all valid quadratic equations, while completing the square is particularly useful if you need exact surd solutions.

**Worked example:** Solve $2x^2 - 3x -7 = 0$, give your answer to 2 decimal places

1. Identify coefficients (including signs): $a=2$, $b=-3$, $c=-7$
2. $$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)} = \frac{3 \pm \sqrt{9 + 56}}{4} = \frac{3 \pm \sqrt{65}}{4}$$
3. Calculate approximate values: $\sqrt{65} \approx 8.062$, so $x \approx \frac{3 + 8.062}{4} = 2.77$ and $x \approx \frac{3 - 8.062}{4} = -1.27$

**Worked example:** Solve $x^2 + 6x + 4 = 0$ by completing the square, leave your answer in exact surd form

1. Rewrite the $x^2 + 6x$ term as a perfect square plus a constant: $(x + 3)^2 - 9$
2. Substitute back into the equation: $(x + 3)^2 - 9 + 4 = 0 \implies (x + 3)^2 = 5$
3. Take square root of both sides (include $\pm$): $x + 3 = \pm \sqrt{5}$
4. Solve for $x$: $x = -3 \pm \sqrt{5}$

> **Exam tip:** If the question asks for exact solutions, do not convert surds to decimals, leave them in root form to avoid losing accuracy marks.

*Calculator:* allowed

## Forming and Solving Contextual Quadratic Equations

You will be given real-world scenarios (e.g. area calculations, number problems, motion) and asked to construct a quadratic equation before solving it. Always check that your solutions make sense in the context of the problem -- for example, length, time, or number of items cannot be negative, so you should discard any negative roots in these cases.

**Worked example:** The length of a rectangle is 3cm longer than its width. The area of the rectangle is $40cm^2$. Find the width of the rectangle.

1. Define variables: let width = $x$ cm, so length = $x + 3$ cm
2. Construct equation from area formula: $x(x + 3) = 40$
3. Rearrange to standard form: $x^2 + 3x - 40 = 0$
4. Factor and solve: $(x + 8)(x - 5) = 0 \implies x = -8$ or $x = 5$
5. Discard invalid negative solution: width = 5cm

> **Exam tip:** Always state clearly which variable you are defining at the start of contextual problems, this makes your working easier to follow for exam markers.

## Solving Linear-Quadratic Simultaneous Equations

For systems with one linear and one quadratic equation, substitute the linear expression for one variable directly into the quadratic equation. This will give you a single quadratic equation in one variable, which you can solve using any of the methods above. You will usually get two values for the first variable; substitute each back into the linear equation to find the corresponding value of the second variable, giving two full solution pairs.

**Worked example:** Solve the simultaneous equations: $y = 2x - 11$ and $x^2 + y^2 = 25$

1. Substitute $y = 2x - 11$ into the quadratic equation: $x^2 + (2x - 11)^2 = 25$
2. Expand and simplify: $x^2 + 4x^2 - 44x + 121 = 25 \implies 5x^2 - 44x + 96 = 0$
3. Factor and solve for $x$: $(5x - 24)(x - 4) = 0 \implies x = 4.8$ or $x = 4$
4. Substitute back to find $y$: for $x=4$, $y=2(4)-11=-3$; for $x=4.8$, $y=2(4.8)-11=-1.4$
5. Final solutions: $(4, -3)$ and $(4.8, -1.4)$

> **Exam tip:** Always substitute both solution pairs back into both original equations to check they work -- arithmetic errors when expanding brackets are very common for this question type.

## Common pitfalls

- **Wrong:** Forgetting to rearrange the quadratic equation to equal 0 before applying solving methods
  - Why it fails: Factorisation, the quadratic formula, and completing the square only work for equations in standard form with RHS = 0, so you will get incorrect roots if you skip this step
  - Correct: Always move all terms to the left-hand side so the right-hand side is 0 before using any quadratic solving method
- **Wrong:** Dropping the $\pm$ sign when using the quadratic formula or completing the square
  - Why it fails: Quadratic equations can have up to two distinct real roots, and the $\pm$ sign accounts for both solutions
  - Correct: Always include the $\pm$ symbol when taking the square root of both sides or applying the quadratic formula
- **Wrong:** Sign errors when identifying $a, b, c$ for the quadratic formula
  - Why it fails: The quadratic formula relies on the correct sign of all coefficients, so even small sign errors lead to completely wrong calculations
  - Correct: Write the standard form equation clearly, circle each coefficient including its sign, before substituting into the formula
- **Wrong:** Keeping invalid negative solutions in contextual problems
  - Why it fails: Real-world quantities like length, time, or number of items cannot be negative, so these solutions are not valid
  - Correct: After solving a contextual quadratic, check each solution against the problem scenario and discard any impossible values
- **Wrong:** Only finding one solution pair for linear-quadratic simultaneous equations
  - Why it fails: The quadratic equation usually has two roots, each corresponding to a full $(x,y)$ solution pair
  - Correct: After finding both $x$-values, substitute each back into the linear equation to find both corresponding $y$-values
- **Wrong:** Converting exact surd/fraction solutions to decimals when not requested
  - Why it fails: Exact solutions are required unless a specific decimal accuracy is stated, so converting them loses accuracy marks
  - Correct: Leave solutions as fractions or surds unless the question explicitly asks for a rounded decimal value

## Cheatsheet

| Method | Suitable For | Key Steps | Tier |
| --- | --- | --- | --- |
| Factorisation ($a=1$) | $x^2 + bx + c = 0$ with integer factors | 1. Rearrange to =0 2. Find factors of $c$ that sum to $b$ 3. Solve for $x$ | Foundation / Higher |
| General Factorisation ($a \neq 1$) | $ax^2 + bx + c =0$ with integer factors | 1. Rearrange to =0 2. Find factors of $a \times c$ that sum to $b$ 3. Factor by grouping | Higher |
| Quadratic Formula | All quadratic equations | 1. Identify $a,b,c$ 2. Substitute into formula 3. Simplify to exact/decimal form | Higher |
| Completing the Square | Quadratics requiring exact surd solutions | 1. Rewrite $x^2 + bx$ as perfect square 2. Isolate squared term 3. Solve for $x$ | Higher |
| Linear-Quadratic Systems | 1 linear + 1 quadratic equation | 1. Substitute linear into quadratic 2. Solve for first variable 3. Find second variable | Higher |

## What's next

Now that you have mastered quadratic equations for Edexcel IGCSE Mathematics A, you are ready to build on this foundational skill for more advanced algebra topics. Quadratic equations appear frequently in later content including quadratic graph sketching, quadratic inequalities (specification point 2.8), and kinematics problems in the applied section of the course. Practise as many exam-style questions as possible, focusing on both accuracy and speed, as quadratics are tested in almost every Higher and Foundation tier paper. Make sure you only revise content relevant to your tier to avoid wasting time on out-of-scope material.

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