Completing the Square

What you'll learn

How to turn expressions with an $x^2$ term into a squared-bracket form.
How to find the lowest or highest point on the graph quickly.
How to handle odd middle numbers and a number in front of $x^2$ .
How to solve equations exactly using square roots.

Key vocabulary

Definition

Essential terms

Completing the square means rewriting an expression like $x^2+px+q$ so it contains a squared bracket.
A quadratic expression has highest power $x^2$ ; its graph is a parabola, a U-shaped curve.
A coefficient is the number multiplying a variable; a constant is a number on its own.
A perfect square is a squared bracket such as $x-3)^2$ .
Completed square form is $x+h)^2+k$ or $a(x+h)^2+k$ , where the letters stand for constants.
A turning point is where the parabola changes direction; it is also called the vertex.

1. Reverse a square bracket

The key expansion is:

x+r)^2=x^2+2rx+r^2

So completing the square is really running this expansion backwards. The middle term, $2 r x$ , tells you that the number in the bracket is half the coefficient of $x$ .

The square of side length x+r splits into x², two xr rectangles, and an r² corner square, showing why the middle term is 2rx.

Key Idea

Half, then square

For $x^2+px+q$ , half the coefficient of $x$ to make the bracket, then subtract the square you accidentally added.

Completing the square adds a missing corner square of side p/2, so its area must be subtracted to keep the expression equal.

Example

Reversing the expansion

The coefficient of $x$ in $x^2+10x$ is 10, so half is 5.
Build the square bracket and expand it:
$x+5)^2=x^2+10x+25$
The square contains 25 too much, so subtract 25:
$x^2+10x=(x+5)^2-25$

2. Completing the square when the $x^2$ coefficient is 1

For an expression in the form $x^2+px+q$ , use:

x2+px+q=(x+p2)2+q−(p2)2x^2+px+q=\left(x+\frac{p}{2}\right)^2+q-\left(\frac{p}{2}\right)^2

You do not need to memorise this as a “big formula” if you remember the half-and-square idea.

Example

Rewrite and find the turning point

Rewrite $x^2-8x+9$ in completed square form, then find the turning point of $y=x^2-8x+9$ .

The completed square form shows the graph shifted right 4 and down 7, with turning point (4, -7).

The coefficient of $x$ is -8, and half of -8 is -4.
Create the bracket $x-4)^2$ , then correct the extra constant:
$x2−8x+9=(x−4)2−16+9=(x−4)2−7\begin{aligned} x^2-8x+9 &=(x-4)^2-16+9\\ &=(x-4)^2-7 \end{aligned}$
So the completed square form is $x-4)^2-7$ .
The square part is smallest when $x = 4$ , so the turning point is $(4, - 7)$ .

Common Mistake

Subtracting the half, not the square

If the bracket is $x-4)^2$ , the extra constant is 16, not 4. You correct by subtracting 16.

3. Reading turning points from completed square form

In $y=(x+h)^2+k$ , the square is always zero or positive. So the lowest value happens when the bracket equals zero.

That means:

point(−h,k)y=(x+h)^2+k \quad \text{has turning point} \quad (-h,k)

If the number multiplying the square is positive, the point is a minimum, meaning the lowest point. If it is negative, it is a maximum, meaning the highest point.

Tip

Opposite sign in the bracket

The x-coordinate uses the opposite sign to the bracket number: $x+6)^2$ gives $x = - 6$ .

A bracket of x+6 becomes zero at x=-6, so the vertex lies 6 units left of the y-axis.

4. Odd coefficients often give fractions

If the coefficient of $x$ is odd, halving it gives a fraction. That is completely normal at higher GCSE.

Example

Completing the square with fractions

Find the turning point of $y=x^2-3x+2$ .

$Halving the odd coefficient gives a fractional vertex at (3/2, -1/4).$

Half of -3 is $−32-\frac{3}{2}$ , so the bracket is $(x−32)2\left(x-\frac{3}{2}\right)^2$ .
Square the half-coefficient:
$(−32)2=94\left(-\frac{3}{2}\right)^2=\frac{9}{4}$
Complete the square:
$x2−3x+2=(x−32)2−94+2=(x−32)2−14\begin{aligned} x^2-3x+2 &=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+2\\ &=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} \end{aligned}$
The turning point is $(32,−14)\left(\frac{3}{2},-\frac{1}{4}\right)$ .

5. When the coefficient of $x^2$ is not 1

The leading coefficient is the coefficient of $x^2$ . If it is not 1, factor it out of the $x^2$ and $x$ terms first, then complete the square inside the bracket.

Example

A coefficient in front of $x^{2}$

Rewrite $2x^2+16x+35$ in the form $a(x+b)^2+c$ , then state the turning point.

The multiplier 2 makes the parabola narrower, while the completed square form gives turning point (-4, 3).

Factor 2 from the first two terms:
$2x^2+16x+35=2(x^2+8x)+35$
Complete the square inside the bracket:
$x^2+8x=(x+4)^2-16$
Substitute this back and remember to multiply the correction by 2:
$2(x2+8x)+35=2((x+4)2−16)+35=2(x+4)2−32+35=2(x+4)2+3\begin{aligned} 2(x^2+8x)+35 &=2\left((x+4)^2-16\right)+35\\ &=2(x+4)^2-32+35\\ &=2(x+4)^2+3 \end{aligned}$
The turning point is $(- 4, 3)$ .

Common Mistake

Forgetting the outside multiplier

If you subtract 16 inside a bracket multiplied by 2, the total change is 32. The outside multiplier affects the correction.

6. Solving by completing the square

To solve a quadratic equation by completing the square:

Complete the square.
Isolate the squared bracket.
Square root both sides.
Use the $±\pm$ sign.

Definition

Surd form

A surd is an exact root that is not a whole number, such as $6\sqrt{6}$ ; surd form means leaving it exact instead of rounding.

Example

Solving exactly with a decimal constant

Solve $x^2+7x+6.25=0$ by completing the square.

The exact solutions are the two x-values where the shifted parabola crosses the x-axis, symmetrically around x=-7/2.

Write 6.25 as an exact fraction:
$6.25=2546.25=\frac{25}{4}$
Half of 7 is $72\frac{7}{2}$ , so complete the square:
$x2+7x+254=(x+72)2−494+254=(x+72)2−6\begin{aligned} x^2+7x+\frac{25}{4} &=\left(x+\frac{7}{2}\right)^2-\frac{49}{4}+\frac{25}{4}\\ &=\left(x+\frac{7}{2}\right)^2-6 \end{aligned}$
Set the completed square expression equal to zero:
$(x+72)2−6=0\left(x+\frac{7}{2}\right)^2-6=0$
Isolate the square:
$(x+72)2=6\left(x+\frac{7}{2}\right)^2=6$
Square root both sides using $±\pm$ , then subtract $72\frac{7}{2}$ : $x=−72±6x=-\frac{7}{2}\pm\sqrt{6}$ .

Common Mistake

Missing one solution

When you square root both sides, remember the $±\pm$ sign. Without it, you only find one of the two roots.

Exam technique

In the exam

Show the halving step clearly, because it earns method marks.
For turning points, read the x-coordinate with the opposite sign to the bracket.
When solving, isolate the squared bracket before square rooting, and leave exact roots in surd form.

Self review

Check yourself

Complete the square for $x^2+12x-5$ .
What is the turning point of $y=2(x-3)^2+7$ ?
Why must you write $±\pm$ when solving $x+1)^2=11$ ?

Completing the Square

What you'll learn

How to turn expressions with an $x^2$ term into a squared-bracket form.
How to find the lowest or highest point on the graph quickly.
How to handle odd middle numbers and a number in front of $x^2$ .
How to solve equations exactly using square roots.

Key vocabulary

Definition

Essential terms

Completing the square means rewriting an expression like $x^2+px+q$ so it contains a squared bracket.
A quadratic expression has highest power $x^2$ ; its graph is a parabola, a U-shaped curve.
A coefficient is the number multiplying a variable; a constant is a number on its own.
A perfect square is a squared bracket such as $x-3)^2$ .
Completed square form is $x+h)^2+k$ or $a(x+h)^2+k$ , where the letters stand for constants.
A turning point is where the parabola changes direction; it is also called the vertex.

1. Reverse a square bracket

The key expansion is:

x+r)^2=x^2+2rx+r^2

So completing the square is really running this expansion backwards. The middle term, $2 r x$ , tells you that the number in the bracket is half the coefficient of $x$ .

The square of side length x+r splits into x², two xr rectangles, and an r² corner square, showing why the middle term is 2rx.

Key Idea

Half, then square

For $x^2+px+q$ , half the coefficient of $x$ to make the bracket, then subtract the square you accidentally added.

Completing the square adds a missing corner square of side p/2, so its area must be subtracted to keep the expression equal.

Example

Reversing the expansion

The coefficient of $x$ in $x^2+10x$ is 10, so half is 5.
Build the square bracket and expand it:
$x+5)^2=x^2+10x+25$
The square contains 25 too much, so subtract 25:
$x^2+10x=(x+5)^2-25$

2. Completing the square when the $x^2$ coefficient is 1

For an expression in the form $x^2+px+q$ , use:

x2+px+q=(x+p2)2+q−(p2)2x^2+px+q=\left(x+\frac{p}{2}\right)^2+q-\left(\frac{p}{2}\right)^2

You do not need to memorise this as a “big formula” if you remember the half-and-square idea.

Example

Rewrite and find the turning point

Rewrite $x^2-8x+9$ in completed square form, then find the turning point of $y=x^2-8x+9$ .

The completed square form shows the graph shifted right 4 and down 7, with turning point (4, -7).

The coefficient of $x$ is -8, and half of -8 is -4.
Create the bracket $x-4)^2$ , then correct the extra constant:
$x2−8x+9=(x−4)2−16+9=(x−4)2−7\begin{aligned} x^2-8x+9 &=(x-4)^2-16+9\\ &=(x-4)^2-7 \end{aligned}$
So the completed square form is $x-4)^2-7$ .
The square part is smallest when $x = 4$ , so the turning point is $(4, - 7)$ .

Common Mistake

Subtracting the half, not the square

If the bracket is $x-4)^2$ , the extra constant is 16, not 4. You correct by subtracting 16.

3. Reading turning points from completed square form

In $y=(x+h)^2+k$ , the square is always zero or positive. So the lowest value happens when the bracket equals zero.

That means:

point(−h,k)y=(x+h)^2+k \quad \text{has turning point} \quad (-h,k)

If the number multiplying the square is positive, the point is a minimum, meaning the lowest point. If it is negative, it is a maximum, meaning the highest point.

Tip

Opposite sign in the bracket

The x-coordinate uses the opposite sign to the bracket number: $x+6)^2$ gives $x = - 6$ .

A bracket of x+6 becomes zero at x=-6, so the vertex lies 6 units left of the y-axis.

4. Odd coefficients often give fractions

If the coefficient of $x$ is odd, halving it gives a fraction. That is completely normal at higher GCSE.

Example

Completing the square with fractions

Find the turning point of $y=x^2-3x+2$ .

$Halving the odd coefficient gives a fractional vertex at (3/2, -1/4).$

Half of -3 is $−32-\frac{3}{2}$ , so the bracket is $(x−32)2\left(x-\frac{3}{2}\right)^2$ .
Square the half-coefficient:
$(−32)2=94\left(-\frac{3}{2}\right)^2=\frac{9}{4}$
Complete the square:
$x2−3x+2=(x−32)2−94+2=(x−32)2−14\begin{aligned} x^2-3x+2 &=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+2\\ &=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} \end{aligned}$
The turning point is $(32,−14)\left(\frac{3}{2},-\frac{1}{4}\right)$ .

5. When the coefficient of $x^2$ is not 1

The leading coefficient is the coefficient of $x^2$ . If it is not 1, factor it out of the $x^2$ and $x$ terms first, then complete the square inside the bracket.

Example

A coefficient in front of $x^{2}$

Rewrite $2x^2+16x+35$ in the form $a(x+b)^2+c$ , then state the turning point.

The multiplier 2 makes the parabola narrower, while the completed square form gives turning point (-4, 3).

Factor 2 from the first two terms:
$2x^2+16x+35=2(x^2+8x)+35$
Complete the square inside the bracket:
$x^2+8x=(x+4)^2-16$
Substitute this back and remember to multiply the correction by 2:
$2(x2+8x)+35=2((x+4)2−16)+35=2(x+4)2−32+35=2(x+4)2+3\begin{aligned} 2(x^2+8x)+35 &=2\left((x+4)^2-16\right)+35\\ &=2(x+4)^2-32+35\\ &=2(x+4)^2+3 \end{aligned}$
The turning point is $(- 4, 3)$ .

Common Mistake

Forgetting the outside multiplier

If you subtract 16 inside a bracket multiplied by 2, the total change is 32. The outside multiplier affects the correction.

6. Solving by completing the square

To solve a quadratic equation by completing the square:

Complete the square.
Isolate the squared bracket.
Square root both sides.
Use the $±\pm$ sign.

Definition

Surd form

A surd is an exact root that is not a whole number, such as $6\sqrt{6}$ ; surd form means leaving it exact instead of rounding.

Example

Solving exactly with a decimal constant

Solve $x^2+7x+6.25=0$ by completing the square.

The exact solutions are the two x-values where the shifted parabola crosses the x-axis, symmetrically around x=-7/2.

Write 6.25 as an exact fraction:
$6.25=2546.25=\frac{25}{4}$
Half of 7 is $72\frac{7}{2}$ , so complete the square:
$x2+7x+254=(x+72)2−494+254=(x+72)2−6\begin{aligned} x^2+7x+\frac{25}{4} &=\left(x+\frac{7}{2}\right)^2-\frac{49}{4}+\frac{25}{4}\\ &=\left(x+\frac{7}{2}\right)^2-6 \end{aligned}$
Set the completed square expression equal to zero:
$(x+72)2−6=0\left(x+\frac{7}{2}\right)^2-6=0$
Isolate the square:
$(x+72)2=6\left(x+\frac{7}{2}\right)^2=6$
Square root both sides using $±\pm$ , then subtract $72\frac{7}{2}$ : $x=−72±6x=-\frac{7}{2}\pm\sqrt{6}$ .

Common Mistake

Missing one solution

When you square root both sides, remember the $±\pm$ sign. Without it, you only find one of the two roots.

Exam technique

In the exam

Show the halving step clearly, because it earns method marks.
For turning points, read the x-coordinate with the opposite sign to the bracket.
When solving, isolate the squared bracket before square rooting, and leave exact roots in surd form.

Self review

Check yourself

Complete the square for $x^2+12x-5$ .
What is the turning point of $y=2(x-3)^2+7$ ?
Why must you write $±\pm$ when solving $x+1)^2=11$ ?

Completing the Square

What you'll learn

Key vocabulary

1. Reverse a square bracket

2. Completing the square when the $x^2$ coefficient is 1

3. Reading turning points from completed square form

4. Odd coefficients often give fractions

5. When the coefficient of $x^2$ is not 1

6. Solving by completing the square

Recap questions

You've reached the end

Completing the Square

What you'll learn

Key vocabulary

1. Reverse a square bracket

2. Completing the square when the $x^2$ coefficient is 1

3. Reading turning points from completed square form

4. Odd coefficients often give fractions

5. When the coefficient of $x^2$ is not 1

6. Solving by completing the square

Recap questions

You've reached the end

Completing the Square

What you'll learn

Key vocabulary

1. Reverse a square bracket

2. Completing the square when the x2x^2x2 coefficient is 1

3. Reading turning points from completed square form

4. Odd coefficients often give fractions

5. When the coefficient of x2x^2x2 is not 1

6. Solving by completing the square

Recap questions

You've reached the end

Completing the Square Revision Guide

Completing the Square

What you'll learn

Key vocabulary

1. Reverse a square bracket

2. Completing the square when the x2x^2x2 coefficient is 1

3. Reading turning points from completed square form

4. Odd coefficients often give fractions

5. When the coefficient of x2x^2x2 is not 1

6. Solving by completing the square

Recap questions

You've reached the end

Completing the Square Revision Guide

2. Completing the square when the $x^2$ coefficient is 1

5. When the coefficient of $x^2$ is not 1

2. Completing the square when the $x^2$ coefficient is 1

5. When the coefficient of $x^2$ is not 1