Common Completing the Square Mistakes at GCSE

A student can understand completing the square and still lose marks on it. The trouble is rarely one enormous misconception. More often, a small error slips into the middle: the wrong number is halved, a square is not applied, or a negative sign disappears.

This guide follows those errors one at a time. The common mistakes with completing the square in GCSE Maths become much easier to avoid once you know where to look. The central method is simple: halve the coefficient of $x$, place that number inside the bracket, then compensate for the extra square you have introduced.

The completing the square checklist

To rewrite

$$x^2+bx+c$$

in completed-square form:

• halve $b$;
• write the result in $(x+\frac{b}{2}{)}^2$;
• subtract $(\frac{b}{2}{)}^2$ to balance the expression;
• include the original constant $c$;
• simplify carefully;
• expand your final answer to check it.

The identity behind the method is

$$x^2+bx+c={\left(x+\frac{b}{2}\right)}^2+c-{\left(\frac{b}{2}\right)}^2.$$

That is worth understanding rather than merely memorising. When the bracket is expanded,

$${\left(x+\frac{b}{2}\right)}^2=x^2+bx+\frac{b^2}{4}.$$

The term $\frac{b^2}{4}$ was not in the original quadratic, so it must be subtracted outside the bracket.

You can review the full method through the Maths Genie completing the square lesson before focusing on the errors below.

Mistake: halving the constant instead of the coefficient

Suppose a student is asked to complete the square for

$$x^2+10x+7.$$

They may look at the final number and write $(x+\frac{7}{2}{)}^2$. This happens because three numbers are competing for attention.

The number to halve is always the coefficient of $x$. Here, that coefficient is $10$, so half of it is $5$:

$$x^2+10x+7=(x+5{)}^2-25+7.$$

Therefore,

$$x^2+10x+7=(x+5{)}^2-18.$$

Expand to check:

$$(x+5{)}^2-18=x^2+10x+25-18=x^2+10x+7.$$

A useful habit is to underline the $x$ term before starting. It gives your attention somewhere precise to land.

Mistake: forgetting to square the halved number

Another student correctly halves $8$ to get $4$, but writes

$$x^2+8x+3=(x+4{)}^2-4+3.$$

The bracket introduces $4^2$, not merely $4$. The compensation must therefore be $-16$:

$$x^2+8x+3=(x+4{)}^2-16+3$$ $$=(x+4{)}^2-13.$$

This is among the most frequent completing the square mistakes because the first instruction is “halve”, while the second operation is easy to overlook. Say the method as two separate actions: halve, then square.

If you want the method beside you as you revise, use the completing the square revision guide.

Mistake: using the wrong sign inside the bracket

Consider

$$x^2-12x+5.$$

Half of $-12$ is $-6$, so the bracket must be $(x-6{)}^2$. Writing $(x+6{)}^2$ would produce a middle term of $+12x$ when expanded.

The correct method is

$$x^2-12x+5=(x-6{)}^2-36+5$$ $$=(x-6{)}^2-31.$$

The sign inside the bracket matches the sign of the $x$ coefficient when the coefficient of $x^2$ is $1$.

A stick-figure student uses a magnifying glass to find a troublesome minus sign

Mistake: mishandling a coefficient greater than one

The method needs an extra first step when the coefficient of $x^2$ is not $1$. For example, complete the square for

$$2x^2+12x+5.$$

A common incorrect start is $(2x+6{)}^2$. That square begins with $4x^2$, so it cannot match the original expression.

Instead, factor $2$ from the first two terms:

$$2x^2+12x+5=2(x^2+6x)+5.$$

Now complete the square inside the bracket:

$$=2\left((x+3{)}^2-9\right)+5.$$

Then multiply everything inside by $2$:

$$=2(x+3{)}^2-18+5$$ $$=2(x+3{)}^2-13.$$

The check confirms it:

$$2(x+3{)}^2-13=2(x^2+6x+9)-13=2x^2+12x+5.$$

The outer coefficient affects both terms inside the large bracket. Forgetting to multiply the correction by that coefficient is another easy way to lose a mark.

Once this sequence feels clear, the Maths Genie completing the square questions let you practise it with worked solutions and mark schemes instead of having to assemble a question set yourself.

Mistake: stopping before simplifying

An answer such as

$$(x+7{)}^2-49+11$$

may show the right method, but it is not fully simplified. Combine the constants:

$$(x+7{)}^2-38.$$

In an exam, a correct intermediate line can earn method marks, but the requested final form still matters. Show the balancing step and then simplify it. Clear working also makes it easier for you to find an error before moving on.

Mistake: misreading the turning point

Completed-square form is often used to identify the turning point of a quadratic. If

$$y=(x-4{)}^2+3,$$

the turning point is $(4,3)$, not $(-4,3)$. The horizontal coordinate has the opposite sign to the one displayed beside $x$ because the square becomes zero when

$$x-4=0,$$

so $x=4$.

Likewise, for

$$y=(x+2{)}^2-7,$$

the turning point is $(-2,-7)$.

Do not rely on a slogan alone. Set the bracket equal to zero. It takes one line and removes the ambiguity.

How to check completed-square form

Expanding is the most reliable check. Your result should reproduce all three parts of the original quadratic: the $x^2$ term, the $x$ term and the constant.

For

$$3(x-2{)}^2+1,$$

expand carefully:

$$3(x^2-4x+4)+1$$ $$=3x^2-12x+12+1$$ $$=3x^2-12x+13.$$

If even one term differs from the original expression, return to the line where you halved the coefficient or balanced the added square. Checking is not an admission that you are unsure. It is part of the method.

Exam-style practice

Try these without looking back at the examples:

• Write $x^2+14x+20$ in completed-square form.
• Write $x^2-6x-4$ in completed-square form.
• Write $3x^2+18x+10$ in completed-square form.
• Find the turning point of $y=x^2+8x+1$.

Answers:

$$x^2+14x+20=(x+7{)}^2-29$$ $$x^2-6x-4=(x-3{)}^2-13$$ $$3x^2+18x+10=3(x+3{)}^2-17$$ $$y=x^2+8x+1=(x+4{)}^2-15,$$

so the turning point is $(-4,-15)$.

After attempting them independently, compare your working with the completing the square worked solutions. Do not check only the final answer. Compare the line where the square was introduced, because that is usually where the decisive mark is won or lost.

Turn corrected mistakes into reliable marks

The student who improves at completing the square does not simply become more careful. They use a repeatable routine: identify the coefficient of $x$, halve it, square it, balance it and expand to check. Knowing the routine is the first step; following it across unfamiliar exam questions is what makes it dependable.

Your turn: start the Maths Genie completing the square Lesson and Question Bank and practise spotting each error before it costs you a mark.