Solving Linear and Quadratic Equations Revision Notes | GCSE AQA Higher Further Maths

1. Why Equations Matter

Every time you find an unknown — a missing length, a break-even price, the time a ball hits the ground — you are solving an equation. Linear equations typically give one answer, though they can have no solution or infinitely many solutions in special cases; quadratic equations can give two, one, or no real solutions.

This lesson covers both types from scratch. By the end you will have four methods for quadratics and know exactly when to reach for each one.

2. Solving Linear Equations

A linear equation has one unknown raised to the power of 1. The strategy is always the same: isolate the unknown by applying inverse operations to both sides, keeping the equation balanced throughout.

Worked example: Solving a Two-Step Linear Equation

Solve $3x - 7 = 14$

1. $3x - 7 + 7 = 14 + 7$ $3x = 21$ — Inverse of subtracting 7 is adding 7. This isolates the term with $x$. [M1]
2. $x = \frac{21}{3} = 7$ — Inverse of multiplying by 3 is dividing by 3. [A1]
3. $3(7) - 7 = 21 - 7 = 14$ ✓ — Substituting $x = 7$ into the original equation confirms the answer is correct.

Answer: $x = 7$

When the unknown appears on both sides, collect all $x$ terms on one side first, then apply inverse operations as normal.

Worked example: Linear Equation with $x$ on Both Sides

Solve $5x + 3 = 2x + 12$

1. $5x - 2x + 3 = 12$ $3x + 3 = 12$ — Collect all $x$ terms on the left. [M1]
2. $3x = 9$ — Remove the constant from the left side.
3. $x = \frac{9}{3} = 3$ — Isolate $x$. [A1]
4. $5(3) + 3 = 18$ and $2(3) + 12 = 18$ ✓ — Both sides equal 18, confirming $x = 3$.

Answer: $x = 3$

3. Introducing Quadratic Equations

Once the power of the unknown rises to 2, you have a quadratic. The standard form is $ax^2 + bx + c = 0$, and there are four methods to solve it: factorisation, reading from a graph, completing the square, and the quadratic formula.

The method you choose depends on the numbers involved. Factorisation is fastest when it works; the formula is the reliable fallback for any quadratic.

4. Method 1 — Solving by Factorisation

Factorisation works by rewriting the quadratic as a product of two brackets. When a product equals zero, at least one factor must be zero — that is the zero-product property, and it is the engine behind this method.

To factorise $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. Those numbers go into the brackets.

Worked example: Solving a Quadratic by Factorisation

Solve $x^2 - 5x + 6 = 0$

1. $-2 \times -3 = 6$ and $-2 + (-3) = -5$ ✓ — Both conditions are satisfied by $-2$ and $-3$. [M1]
2. $(x - 2)(x - 3) = 0$ — Check by expanding: $x^2 - 3x - 2x + 6 = x^2 - 5x + 6$ ✓ [M1]
3. $x - 2 = 0 \Rightarrow x = 2$ $x - 3 = 0 \Rightarrow x = 3$ — Each bracket gives one solution. [A1 each]

Answer: $x = 2$ or $x = 3$

When $a eq 1$, the factorisation is slightly more involved. You need two numbers that multiply to $ac$ and add to $b$, then split the middle term.

Worked example: Factorising When $a \neq 1$

Solve $2x^2 + 5x - 3 = 0$

1. $6 \times -1 = -6$ and $6 + (-1) = 5$ ✓ — The two numbers are $6$ and $-1$. [M1]
2. $2x^2 + 6x - x - 3 = 0$ — Replace $5x$ with $6x - x$.
3. $2x(x + 3) - 1(x + 3) = 0$ $(2x - 1)(x + 3) = 0$ — Both groups share the factor $(x + 3)$. [M1]
4. $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ $x + 3 = 0 \Rightarrow x = -3$ — [A1 each]

Answer: $x = \frac{1}{2}$ or $x = -3$

5. Method 2 — Solving by Graph

The solutions of $ax^2 + bx + c = 0$ are the $x$-coordinates where the parabola $y = ax^2 + bx + c$ crosses the $x$-axis. This is because at those points $y = 0$, which is exactly the equation you are solving.

The parabola above crosses the $x$-axis at $x = 2$ and $x = 3$, confirming the factorisation result from the previous section. When a question asks you to solve by graph, read the $x$-intercepts carefully — these are your solutions.

A parabola that touches but does not cross the $x$-axis has one repeated root; one that sits entirely above or below the axis has no real roots. The graph makes this visual and immediate.

6. Method 3 — Completing the Square

Completing the square is a powerful technique that rewrites any quadratic in the form $(x + p)^2 + q = 0$. From there, you isolate the bracket and take the square root of both sides.

The key move: halve the coefficient of $x$, square it, add and subtract it inside the expression. This creates a perfect square trinomial without changing the value of the expression.

Worked example: Solving by Completing the Square

Solve $x^2 + 6x + 7 = 0$ by completing the square. Give your answers in surd form.

1. Coefficient of $x$ is $6$, so half is $3$. — This value goes inside the bracket.
2. $(x + 3)^2 - 3^2 + 7 = 0$ $(x + 3)^2 - 9 + 7 = 0$ $(x + 3)^2 - 2 = 0$ — Subtract $3^2 = 9$ to compensate for adding it inside the bracket. [M1]
3. $(x + 3)^2 = 2$ — Add 2 to both sides.
4. $x + 3 = \pm\sqrt{2}$ — Remember the $\pm$ — both the positive and negative root are valid. [M1]
5. $x = -3 + \sqrt{2}$ or $x = -3 - \sqrt{2}$ — Subtract 3 from both sides. Leave in surd form unless told to round. [A1 each]

Answer: $x = -3 + \sqrt{2}$ or $x = -3 - \sqrt{2}$

When $a eq 1$, divide every term by $a$ first to make the leading coefficient 1, then complete the square as normal.

7. Method 4 — The Quadratic Formula

When factorisation does not work cleanly and completing the square feels slow, the quadratic formula is your reliable tool. It works for every quadratic — no exceptions.

Worked example: Solving with the Quadratic Formula

Solve $2x^2 - 3x - 4 = 0$. Give your answers to 2 decimal places.

1. $a = 2,\quad b = -3,\quad c = -4$ — Read coefficients carefully, including signs. [M1]
2. $b^2 - 4ac = (-3)^2 - 4(2)(-4)$ $= (-3) \times (-3) + 32$ $= 9 + 32 = 41$ — $(-3)^2 = (-3) \times (-3) = 9$. Note $-4ac = -4 \times 2 \times (-4) = +32$. [M1]
3. $x = \frac{-(-3) \pm \sqrt{41}}{2 \times 2} = \frac{3 \pm \sqrt{41}}{4}$ — $-b = -(-3) = 3$. Keep $\sqrt{41}$ exact until the final step.
4. $x = \frac{3 + \sqrt{41}}{4} \approx \frac{3 + 6.403}{4} \approx \frac{9.403}{4} \approx 2.35$ $x = \frac{3 - \sqrt{41}}{4} \approx \frac{3 - 6.403}{4} \approx \frac{-3.403}{4} \approx -0.85$ — $\sqrt{41} \approx 6.403$ (calculator). Round only at the final step. [A1 each]

Answer: $x \approx 2.35$ or $x \approx -0.85$ (to 2 d.p.)

The expression under the square root, $b^2 - 4ac$, is called the discriminant. It tells you how many real solutions exist before you do any further work.

8. Choosing the Right Method

Knowing four methods is only useful if you can pick the right one quickly. The table below gives a decision guide based on what the question looks like.

9. Deliberate Practice

Work through these problems using the method indicated. Check your answers against the solutions provided — pay attention to where signs can trip you up.

10. Exam Technique and Common Mistakes

The methods are straightforward once practised, but there are predictable places where marks are dropped. The guidance below addresses the most frequent errors seen in exam scripts.

That final part illustrates a key exam skill: always check whether both solutions make sense in context. A negative length is mathematically valid but physically impossible, so you must reject it and state why.

You now have the full toolkit for solving linear and quadratic equations — inverse operations for linear equations, and factorisation, graphical reading, completing the square, and the quadratic formula for quadratics. Practise selecting the method before you start, and always verify your solutions by substituting back into the original equation.