Introduction to Completing the Square Revision Notes | GCSE AQA Higher Further Maths

1. Why Completing the Square Matters

You already know how to factorise quadratics — but what happens when a quadratic refuses to factorise neatly? That is where completing the square steps in. It is a method that works on every quadratic, not just the tidy ones.

Beyond solving equations, completing the square unlocks the geometry of a parabola. Once a quadratic is in the form $(x+p)^2 + q$, you can read off the vertex, the line of symmetry, and the minimum or maximum value — all without plotting a single point.

Examiners use this technique across several question types: solving quadratics, sketching graphs, and proving results. Mastering one method gives you access to all three.

2. The Key Idea: Perfect Squares

The whole method rests on one pattern. Expand $(x + 3)^2$ and you get $x^2 + 6x + 9$. Notice that the constant term, $9$, is exactly $3^2$ — the square of half the coefficient of $x$. That relationship is the engine of completing the square.

Think of it like this: you are trying to build a perfect square out of $x^2 + bx$, but it is missing a corner piece. That missing piece is $\left(\frac{b}{2}\right)^2$. You add it to complete the square, then immediately subtract it so the overall value does not change.

3. Completing the Square: Monic Quadratics

A monic quadratic has a leading coefficient of 1, such as $x^2 + 6x + 7$. These are the most straightforward cases, so start here before tackling non-monic examples.

Worked example: Completing the Square: $x^2 + 6x + 7$

Write $x^2 + 6x + 7$ in the form $(x + p)^2 + q$.

1. $\frac{6}{2} = 3$ — Half of 6 is 3. This is the value that goes inside the bracket. [Method — identifying $p$]
2. $(x + 3)^2$ — This expands to $x^2 + 6x + 9$, which matches the $x^2 + 6x$ part of the original — but adds an unwanted $+9$.
3. $(x + 3)^2 - 9$ — Subtracting 9 cancels the extra term introduced in Step 2, keeping the expression equivalent to the original. [M1 for correct bracket and subtraction]
4. $(x + 3)^2 - 9 + 7$ — Now include the $+7$ from the original expression.
5. $(x + 3)^2 - 2$ — $-9 + 7 = -2$. This is the completed square form. [A1 for correct final form]

Answer: $(x + 3)^2 - 2$

From this result, you can read off the vertex immediately: the minimum point is at $(-3, -2)$, and the line of symmetry is $x = -3$. No further working is needed — the form does the work for you.

4. Completing the Square with a Negative $x$-Term

The method is identical when the coefficient of $x$ is negative — you still halve it and square it. The sign inside the bracket changes, but the arithmetic follows the same steps.

Worked example: Completing the Square: $x^2 - 8x + 3$

Write $x^2 - 8x + 3$ in completed square form.

1. $\frac{-8}{2} = -4$ — Half of $-8$ is $-4$. This goes inside the bracket.
2. $(x - 4)^2$ — Expanding gives $x^2 - 8x + 16$, which introduces an unwanted $+16$.
3. $(x - 4)^2 - 16$ — Subtracting 16 restores equivalence. [M1]
4. $(x - 4)^2 - 16 + 3$ — Include the $+3$ from the original expression.
5. $(x - 4)^2 - 13$ — $-16 + 3 = -13$. [A1]

Answer: $(x - 4)^2 - 13$

5. Solving Quadratic Equations by Completing the Square

Once a quadratic is in completed square form, solving it is a matter of isolating $x$ step by step. This approach is especially useful on non-calculator papers, where the quadratic formula would require messy arithmetic.

Worked example: Solving by Completing the Square: $x^2 + 6x + 7 = 0$

Solve $x^2 + 6x + 7 = 0$, giving your answers in surd form. (Non-calculator method)

1. $(x + 3)^2 - 9 + 7 = 0$ — From Section 3, we already know $x^2 + 6x + 7 = (x+3)^2 - 2$. [M1 for completing the square correctly]
2. $(x + 3)^2 - 2 = 0$ — $-9 + 7 = -2$.
3. $(x + 3)^2 = 2$ — Add 2 to both sides. [M1 for correct rearrangement]
4. $x + 3 = \pm\sqrt{2}$ — Remember the $\pm$ — there are two roots. Forgetting the negative root is one of the most common errors on this topic.
5. $x = -3 + \sqrt{2}$ or $x = -3 - \sqrt{2}$ — Subtract 3 from both sides. Leave in surd form unless told to round. [A1 for both correct solutions]

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

On a non-calculator paper, leave the answer in surd form. On a calculator paper, if the question asks for a decimal, compute $\sqrt{2} \approx 1.414$, giving $x \approx -1.586$ or $x \approx -4.414$ — but always show the surd step first to secure method marks.

6. Non-Monic Quadratics: Coefficient of $x^2 \neq 1$

When the coefficient of $x^2$ is not 1, there is one extra step: factor it out from the $x^2$ and $x$ terms before completing the square. The constant term stays outside the factorisation for now.

Worked example: Completing the Square: $2x^2 + 12x + 5$

Write $2x^2 + 12x + 5$ in the form $a(x + p)^2 + q$.

1. $2(x^2 + 6x) + 5$ — Factor 2 from $2x^2 + 12x$. The constant $+5$ stays outside. [M1 for correct factorisation]
2. $2\left[(x + 3)^2 - 9\right] + 5$ — Half of 6 is 3, so the bracket is $(x+3)^2$. Expanding $(x+3)^2$ gives $x^2 + 6x + 9$, so we subtract 9 inside the bracket to compensate.
3. $2(x + 3)^2 - 18 + 5$ — Multiply the $-9$ inside the bracket by the outer factor of 2: $2 \times (-9) = -18$. [M1 for correctly expanding the outer factor]
4. $2(x + 3)^2 - 13$ — $-18 + 5 = -13$. [A1 for correct final form]

Answer: $2(x + 3)^2 - 13$

The vertex of $y = 2x^2 + 12x + 5$ is at $(-3, -13)$, and the line of symmetry is $x = -3$. The coefficient $a = 2$ tells you the parabola opens upwards and is narrower than $y = x^2$, but it does not affect the vertex coordinates.

7. Deliberate Practice

These questions cover the full range of completing the square skills. Work through each one before checking the solution — the method marks come from showing clear, structured working.

8. Exam Technique and Common Mistakes

Completing the square is a reliable source of marks — but only if the method is applied consistently. The errors below appear repeatedly in mark schemes, and each one costs marks that are otherwise straightforward to earn.

Completing the square is one of the most versatile algebraic tools at GCSE. It converts any quadratic into a form that reveals the vertex and line of symmetry directly, enables solving without the quadratic formula, and provides the foundation for sketching parabolas accurately — all from a single, repeatable method.