GCSE Maths · Algebra

Completing the Square GCSE Questions and Answers

To complete the square, halve the coefficient of x to build the squared bracket, then correct the constant term. The completed form reveals a quadratic graph’s turning point and can be used to solve equations exactly.

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Build the nearest perfect square

What you need to know about completing the square

An expression containing x², such as x² + 8x + 3, is a quadratic expression. Completing the square rewrites it as one squared bracket plus or minus a number without changing its value. This form makes the turning point visible and can also help solve an equation.

Quadratic → square → adjustment

See why it is called completing the square

A square with side x + 4 has area (x + 4)². Its pieces are x², two strips of area 4x and a 4 by 4 corner, so its total area is x² + 8x + 16.

Build the square(x + 4)²the two 4x strips make the 8x term

Compare+16 instead of +3the square contains 13 too much

Correct it(x + 4)² − 13this equals x² + 8x + 3

The two 4x strips fit around x². A 4 by 4 corner completes the larger square, so x² + 8x + 16 = (x + 4)².

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Rewrite x² + bx + c

What the problem asks: Write a quadratic whose x² coefficient is 1 as one squared bracket plus or minus a constant.

How to solve it: The number inside the bracket is half the x-coefficient because it appears twice when the bracket is squared. Expand, compare constants and correct the difference.

$x^2+bx+c=\left(x+\frac b2\right)^2+c-\left(\frac b2\right)^2$

Rewrite ax² + bx + c

What the problem asks: Write a quadratic whose x² coefficient is not 1 in completed-square form.

How to solve it: Factor the leading coefficient from the x² and x terms first. Complete the square inside the bracket, then apply the outside multiplier to the adjustment.

Find a turning point

What the problem asks: The graph is written or can be written as y = a(x − p)² + q.

How to solve it: A square is smallest at zero, so the turning point is (p, q). The sign inside the bracket is opposite to the x-coordinate.

$y=a(x-p)^2+q\quad\Rightarrow\quad\text{turning point }(p,q)$

Solve a quadratic exactly

What the problem asks: The equation does not factorise neatly or asks for exact answers.

How to solve it: Complete the square, isolate the squared bracket, then take both the positive and negative square roots.

A reliable routine

Method for quadratics of the form x² + bx + c

This routine applies when you are rewriting x² + bx + c and the coefficient of x² is 1. Halving b works because squaring (x + p) creates two px terms, giving 2px.

Halve the coefficient of x and put that number in a squared bracket with x.
Expand the squared bracket mentally or on a separate line to find its constant term.
Add or subtract outside the bracket so the full expression matches the original.
Expand your final form to check every term, especially the sign of the x-term.

Check: Expanding the completed form must reproduce the original expression exactly. The quadratic formula comes from completing ax² + bx + c = 0 in general; completed-square form is also useful because it reveals the turning point and proves bounds.

Fully worked

Completing the square GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Complete the square with a positive x-term

2 marks

Question

Write $x^2+8x+3$ in the form $(x+a)^2+b$ .

Compare $x^2+8x$ with $(x+p)^2=x^2+2px+p^2$ .

$2p=8$

$p=4$

So use $(x+4)^2$ .

$(x+4)^2=x^2+8x+16$

To change $+16$ to $+3$ , subtract $13$ .

$\boxed{x^2+8x+3=(x+4)^2-13}$

Exam tip: expand the bracket to check the adjustment.

Example 2

Complete the square with a negative x-term

2 marks

Question

Write $x^2-10x+7$ in completed-square form.

Half of $-10$ is $-5$ .

$(x-5)^2=x^2-10x+25$

The required constant is $7$ , which is $18$ less than $25$ .

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

Example 3

Complete the square when the leading coefficient is not 1

3 marks

Question

Write $2x^2+12x-5$ in the form $a(x+b)^2+c$ .

Factor $2$ from the terms containing $x$ .

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

Complete the square inside the bracket.

$=2\big((x+3)^2-9\big)-5$

Multiply the adjustment by $2$ .

$=2(x+3)^2-18-5$

$\boxed{2(x+3)^2-23}$

Example 4

Find a minimum point

3 marks

Question

Find the coordinates of the turning point of $y=x^2-6x+11$ . State whether it is a maximum or minimum.

Complete the square.

$y=(x-3)^2-9+11$

$y=(x-3)^2+2$

The square is smallest when $x-3=0$ , so $x=3$ and $y=2$ .

$\boxed{\text{Minimum point }(3,2)}$

Example 5

Solve exactly by completing the square

4 marks

Question

Solve $x^2+4x-1=0$ . Give exact answers.

Complete the square on the left.

$(x+2)^2-5=0$

$(x+2)^2=5$

Take both square roots.

$x+2=\pm\sqrt5$

$\boxed{x=-2\pm\sqrt5}$

Example 6

Find a maximum with a negative leading coefficient

4 marks

Question

Write $-2x^2+8x+1$ in completed-square form and hence find its maximum value.

Factor $-2$ from the $x$ terms.

$-2x^2+8x+1=-2(x^2-4x)+1$

Complete the square inside.

$=-2\big((x-2)^2-4\big)+1$

$=-2(x-2)^2+8+1$

$=-2(x-2)^2+9$

Because $-2(x-2)^2$ is never positive, its greatest value is $0$ .

$\boxed{\text{Maximum value }9\text{ at }x=2}$

Example 7

Use completed form to sketch a graph

4 marks

Question

Write $y=x^2-4x-5$ in completed-square form. Hence state the turning point, axis of symmetry and y-intercept needed for a sketch.

Complete the square.

$y=(x-2)^2-4-5$

$y=(x-2)^2-9$

The square is smallest when $x=2$ , so the turning point is $(2,-9)$ and the axis of symmetry is $x=2$ .

At the y-axis, $x=0$ , so $y=-5$ .

$\boxed{\text{turning point }(2,-9),\quad x=2,\quad \text{y-intercept }(0,-5)}$

Exam tip: label the turning point and intercept on the sketch; do not rely on its scale.

Example 8

Prove a quadratic is always positive

3 marks

Question

Show that $x^2+4x+7>0$ for every real value of $x$ .

Complete the square.

$x^2+4x+7=(x+2)^2-4+7$

$=(x+2)^2+3$

A square is always at least zero.

$(x+2)^2+3\ge3$

Therefore $\boxed{x^2+4x+7>0}$ for every real $x$ .

15 original questions · total 53 marks

Completing the square GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 60 minutes · show the completed-square form before using it · answers start collapsed

Form a perfect square

2 marks

Write $x^2+14x+49$ as a squared bracket.

Show worked answer

Half of $14$ is $7$ .

$\boxed{x^2+14x+49=(x+7)^2}$

Complete a positive square

2 marks

Write $x^2+12x+8$ in completed-square form.

Show worked answer

$(x+6)^2=x^2+12x+36$

$\boxed{x^2+12x+8=(x+6)^2-28}$

Complete a square with a negative term

3 marks

Write $x^2-7x+4$ in the form $(x+a)^2+b$ .

Show worked answer

Half of $-7$ is $-\frac72$ .

$x^2-7x+4=\left(x-\frac72\right)^2-\frac{49}{4}+4$

$=\left(x-\frac72\right)^2-\frac{33}{4}$

$\boxed{a=-\frac72,\quad b=-\frac{33}{4}}$

Use an outside coefficient

3 marks

Write $3x^2+18x+5$ in the form $a(x+b)^2+c$ .

Show worked answer

$3x^2+18x+5=3(x^2+6x)+5$

$=3\big((x+3)^2-9\big)+5$

$\boxed{3(x+3)^2-22}$

Read a turning point

3 marks

Find the turning point of $y=x^2+10x+21$ .

Show worked answer

$y=(x+5)^2-25+21$

$y=(x+5)^2-4$

The square is zero when $x=-5$ .

$\boxed{(-5,-4)}$

Find a maximum point

4 marks

Find the turning point of $y=-x^2-6x+2$ and state its type.

Show worked answer

$y=-(x^2+6x)+2$

$y=-\big((x+3)^2-9\big)+2$

$y=-(x+3)^2+11$

$\boxed{\text{Maximum point }(-3,11)}$

Solve with integer roots

4 marks

Solve $x^2-8x+7=0$ by completing the square.

Show worked answer

$(x-4)^2-9=0$

$(x-4)^2=9$

$x-4=\pm3$

$\boxed{x=1\text{ or }x=7}$

Solve with surd roots

4 marks

Solve $x^2-2x-6=0$ . Give exact answers.

Show worked answer

$(x-1)^2-7=0$

$(x-1)^2=7$

$x-1=\pm\sqrt7$

$\boxed{x=1\pm\sqrt7}$

Explain a minimum bound

4 marks

The function is $f(x)=2x^2-16x+35$ . Find its minimum value and explain why $f(x)\ge3$ for every real value of $x$ .

Show worked answer

$f(x)=2(x^2-8x)+35$

$=2\big((x-4)^2-16\big)+35$

$=2(x-4)^2+3$

The squared term is at least zero.

Because $(x-4)^2\ge0$ , multiplying it by $2$ cannot make it negative. Therefore

$\boxed{\text{minimum value }3\text{ and }f(x)\ge3}$

Solve a quadratic with a leading coefficient

5 marks

Solve $2x^2+8x-3=0$ by completing the square. Give exact answers.

Show worked answer

Divide the equation by $2$ .

$x^2+4x-\frac32=0$

$(x+2)^2-4-\frac32=0$

$(x+2)^2=\frac{11}{2}$

$x+2=\pm\sqrt{\frac{11}{2}}$

$\boxed{x=-2\pm\frac{\sqrt{22}}{2}}$

Sketch from completed-square form

4 marks

Write $y=x^2+6x+5$ in completed-square form. Hence sketch the graph, showing the turning point and y-intercept.

Show worked answer

$y=(x+3)^2-9+5$

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

The turning point is $(-3,-4)$ . The coefficient of the square is positive, so the curve opens upwards.

At $x=0$ , $y=5$ , so the y-intercept is $(0,5)$ .

$\boxed{\text{turning point }(-3,-4),\quad \text{y-intercept }(0,5)}$

The positive x² coefficient makes the parabola open upwards. The axis of symmetry is x = −3.

Prove a quadratic is positive

3 marks

Show that $x^2-10x+29>0$ for all real $x$ .

Show worked answer

$x^2-10x+29=(x-5)^2-25+29$

$=(x-5)^2+4$

Since $(x-5)^2\ge0$ ,

$x^2-10x+29\ge4$

so $\boxed{x^2-10x+29>0}$ for all real $x$ .

Find intersections with a line

5 marks

The curve $y=x^2-6x+10$ meets the line $y=2$ . Find the x-coordinates of the points of intersection by completing the square.

Show worked answer

Set the two expressions for $y$ equal.

$x^2-6x+10=2$

Complete the square.

$(x-3)^2+1=2$

$(x-3)^2=1$

$x-3=\pm1$

$\boxed{x=2\text{ or }x=4}$

Use a negative leading coefficient

4 marks

Write $-3x^2-12x+1$ in completed-square form and state its maximum value.

Show worked answer

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

$=-3\big((x+2)^2-4\big)+1$

$=-3(x+2)^2+13$

Because $-3(x+2)^2\le0$ ,

$\boxed{\text{maximum value }13}$

Complete the square with a factor

3 marks

Write $4x^2-20x+7$ in the form $a(x-b)^2+c$ .

Show worked answer

Factor $4$ from the terms containing $x$ .

$4x^2-20x+7=4(x^2-5x)+7$

$=4\left(\left(x-\frac52\right)^2-\frac{25}{4}\right)+7$

Multiply the whole adjustment by $4$ .

$=4\left(x-\frac52\right)^2-25+7$

$\boxed{4\left(x-\frac52\right)^2-18}$

Examiner-style feedback

Common completing the square mistakes

Using the whole x-coefficient

The number inside the bracket is half the coefficient of x, not the full coefficient.

Forgetting the outside adjustment

A squared bracket adds its own constant. Correct it outside so the expression remains equivalent.

Not multiplying the adjustment by a

In 2((x + 3)² − 9) − 5, the −9 is inside the bracket, so it also multiplies by 2: the result is 2(x + 3)² − 18 − 5.

Dividing only part of an equation

If you divide 2x² + 8x − 3 = 0 by 2, divide every term, including −3. The new constant is −3/2.

Reading the x-coordinate with the wrong sign

The turning point of (x − p)² + q is (p, q), so the bracket sign appears reversed.

Dropping one solution

When a squared expression equals a positive number, take both the positive and negative square roots.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

Halve the coefficient of x to build the squared bracket.
Correct the constant outside the bracket.
Use the completed form to read a turning point or isolate the square.
Expand to check and keep ± when solving.

Quick answers

Completing the square FAQ

Is completing the square Higher tier?

Yes. Completing the square is Higher-tier GCSE Maths content in the specifications covered by this guide.

Why do you halve the coefficient of x?

Expanding (x + p)² gives x² + 2px + p², so p must be half the coefficient of x.

How does completed-square form give the turning point?

In a(x − p)² + q, the squared part is zero at x = p, giving y = q.

When should I use completing the square to solve?

Use it when the question requests the method, asks for exact solutions, or connects the equation to a graph or turning point.

Build connected skills

What to revise next

Exact answers

Surds

Simplify the exact roots produced by quadratic equations.

Revise surds

Linked algebra

Simultaneous equations

Use quadratics inside linear–quadratic systems.

Revise simultaneous equations

Function language

Functions

Connect completed-square form to quadratic functions.

Revise functions

Content standards

Curriculum and rights review

Curriculum references checked 4 September 2026. Completing the square, solving quadratics and interpreting quadratic turning points are shared Higher-tier GCSE Mathematics content. All questions, values and solution wording are original Pass an Exam material.

Official specification references