Completing the Square and the Quadratic Formula Revision Notes | GCSE AQA Higher Maths

1. Why Factorisation Is Not Always Enough

You already know how to solve quadratics by factorisation — find two numbers that multiply to $c$ and add to $b$, write two brackets, set each to zero. That works beautifully when the numbers cooperate. But what happens when they don't?

Try factorising $x^2 + 4x + 1 = 0$. You need two integers that multiply to 1 and add to 4. There are none. Factorisation fails here, and that is exactly where completing the square and the quadratic formula take over.

Both methods work on any quadratic — not just the tidy ones. Mastering them means no quadratic equation can stop you in an exam, regardless of how awkward the numbers look.

2. Completing the Square

The idea behind completing the square is to rewrite $x^2 + bx$ as a perfect square bracket minus a correction term. Think of it as forcing the expression into a shape you can work with.

The key move is always the same: halve the coefficient of $x$, write it inside a squared bracket, then subtract the square of that half. Once the expression is in completed-square form, solving for $x$ is straightforward — rearrange to isolate $x$.

Worked example: Solving a Quadratic by Completing the Square

Solve $x^2 + 6x + 1 = 0$, giving your answers in surd form. [3 marks]

1. $\frac{6}{2} = 3$ — We will write $(x + 3)^2$ as the squared bracket. [M1 — method for completing the square]
2. $(x + 3)^2 - 3^2 + 1 = 0$ $(x + 3)^2 - 9 + 1 = 0$ $(x + 3)^2 - 8 = 0$ — Subtract $3^2 = 3 \times 3 = 9$ to compensate for the extra term introduced by squaring the bracket, then simplify the constants.
3. $(x + 3)^2 = 8$ — Add 8 to both sides. The equation is now in a form we can square-root.
4. $x + 3 = \pm\sqrt{8}$ — Remember the $\pm$ — there are two roots. $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$, so $x + 3 = \pm 2\sqrt{2}$.
5. $x = -3 + 2\sqrt{2}$ or $x = -3 - 2\sqrt{2}$ — Subtract 3 from both sides. These are exact surd answers. [A1 for each correct root]

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

Notice that $\sqrt{8}$ was simplified to $2\sqrt{2}$ before writing the final answer. Leaving $\sqrt{8}$ unsimplified in a final answer will cost you the accuracy mark on GCSE Higher exam papers — always simplify surds.

2a. Completing the Square When the Leading Coefficient Is Not 1

When the coefficient of $x^2$ is not 1, factorise it out first before completing the square inside the bracket. This keeps the method consistent and avoids sign errors.

Worked example: Completing the Square with a Leading Coefficient

Write $2x^2 - 8x + 3$ in the form $a(x + p)^2 + q$. [3 marks]

1. $2(x^2 - 4x) + 3$ — Only factorise from the $x^2$ and $x$ terms, not the constant. [M1]
2. $2\left[(x - 2)^2 - 4\right] + 3$ — Halve $-4$ to get $-2$. Then $(-2)^2 = (-2) \times (-2) = 4$, which is subtracted inside the bracket.
3. $2(x - 2)^2 - 8 + 3$ — Multiply the $-4$ inside the bracket by the outer factor of 2: $2 \times (-4) = -8$.
4. $2(x - 2)^2 - 5$ — $-8 + 3 = -5$. The expression is now in the required form with $a = 2$, $p = -2$, $q = -5$. [A1 for correct $p$, A1 for correct $q$]

Answer: $2(x - 2)^2 - 5$

3. The Quadratic Formula

Completing the square is powerful, but it can be slow when the numbers are messy. The quadratic formula is derived directly from completing the square on the general form $ax^2 + bx + c = 0$ — so it works for every quadratic, every time.

Before substituting, always rearrange the equation into the form $ax^2 + bx + c = 0$ — every term on one side, zero on the other. Getting $a$, $b$, and $c$ wrong is the most common source of errors with this formula.

Worked example: Solving a Quadratic Using the Quadratic Formula

Solve $3x^2 + 5x - 2 = 0$, giving your answers to 2 decimal places where necessary. [3 marks]

1. $a = 3,\quad b = 5,\quad c = -2$ — The equation is already in the form $ax^2 + bx + c = 0$. Note that $c = -2$, not $+2$. [M1 — correct identification]
2. $b^2 - 4ac = 5^2 - 4 \times 3 \times (-2)$ $= 5 \times 5 - 4 \times 3 \times (-2)$ $= 25 + 24 = 49$ — Multiplying $4 \times 3 \times (-2) = -24$, and subtracting a negative gives $+24$. The discriminant is positive, confirming two real roots.
3. $x = \frac{-5 \pm \sqrt{49}}{2 \times 3} = \frac{-5 \pm 7}{6}$ — $\sqrt{49} = 7$. This is a calculator-allowed exam question method, though this particular example works out neatly. [M1 — correct substitution]
4. $x = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}$ $x = \frac{-5 - 7}{6} = \frac{-12}{6} = -2$ — Work out the $+$ case and the $-$ case separately. Both answers are exact here, so no rounding is needed. [A1 for both correct solutions]

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

That example factorised neatly in the end — but the formula still worked. Now look at a case where the formula is genuinely necessary because the roots are irrational.

Worked example: Quadratic Formula with Surd Answers

Solve $2x^2 - 3x - 4 = 0$, giving your answers to 2 decimal places. [3 marks]

1. $a = 2,\quad b = -3,\quad c = -4$ — Be careful: $b = -3$, not $+3$. Sign errors here cascade through every subsequent step.
2. $b^2 - 4ac = (-3)^2 - 4 \times 2 \times (-4)$ $= (-3) \times (-3) - 4 \times 2 \times (-4)$ $= 9 + 32 = 41$ — $(-3)^2 = 9$ and $4 \times 2 \times (-4) = -32$, so subtracting $-32$ gives $+32$. [M1]
3. $x = \frac{-(-3) \pm \sqrt{41}}{2 \times 2} = \frac{3 \pm \sqrt{41}}{4}$ — $-b = -(-3) = 3$. Do not round $\sqrt{41}$ yet — keep it exact until the final step.
4. $\sqrt{41} \approx 6.4031...$ $x = \frac{3 + 6.4031}{4} = \frac{9.4031}{4} \approx 2.35$ $x = \frac{3 - 6.4031}{4} = \frac{-3.4031}{4} \approx -0.85$ — Use a calculator for $\sqrt{41}$. Round only at this final step. [A1 for each root to 2 d.p.]

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

The graph shows a U-shaped parabola crossing the $x$-axis at approximately $x = -0.85$ and $x = 2.35$ — exactly the roots found by the formula. Whenever you solve a quadratic, the roots are the $x$-intercepts of the corresponding parabola.

4. The Discriminant and the Number of Roots

You have already been calculating $b^2 - 4ac$ as part of the quadratic formula. This expression has a name — the discriminant — and it tells you how many real solutions a quadratic has before you finish the calculation.

5. Choosing the Right Method

Three methods are available for solving quadratics: factorisation, completing the square, and the quadratic formula. Picking the right one saves time and reduces errors.

In questions without a calculator, completing the square is your primary tool for quadratics that do not factorise. On calculator-allowed exam questions, the quadratic formula is faster for messy coefficients — but you must still show your substitution clearly to earn method marks.

6. Equations That Need Rearranging First

GCSE Higher exam questions rarely present a quadratic already in the form $ax^2 + bx + c = 0$. You will often need to expand brackets, collect like terms, or move everything to one side before applying any method.

Worked example: Solving a Quadratic That Requires Rearrangement

Solve $x(x + 3) = 10$, giving your answers to 2 decimal places. [4 marks]

1. $x^2 + 3x = 10$ — Distribute $x$ across the bracket. [M1 — correct expansion]
2. $x^2 + 3x - 10 = 0$ — Subtract 10 from both sides. Now identify $a = 1$, $b = 3$, $c = -10$.
3. Need two numbers: product $= -10$, sum $= 3$. Try $5$ and $-2$: $5 \times (-2) = -10$ and $5 + (-2) = 3$. Yes. — $(x + 5)(x - 2) = 0$ works here. If it had not factorised, we would use the formula.
4. $x + 5 = 0 \Rightarrow x = -5$ $x - 2 = 0 \Rightarrow x = 2$ — [A1 for each correct root. Both are exact, so no rounding needed despite the question's instruction.]

Answer: $x = 2$ or $x = -5$

7. Deliberate Practice

These questions cover the full range of methods. For each one, decide which method to use before you start — that decision is itself an exam skill.

8. Exam Technique and Common Mistakes

The methods are learnable — but marks are lost through avoidable errors. The mistakes below appear repeatedly in mark schemes, and knowing them in advance is a genuine advantage.

Completing the square and the quadratic formula are the two Higher-tier methods that unlock every quadratic equation — whether the roots are integers, fractions, or surds. Practise identifying $a$, $b$, and $c$ carefully, simplify all surds in final answers, and always show the rearrangement step. Those habits alone will protect your marks across both calculator-allowed questions and questions without a calculator.