Completing the Square

Completing the square rewrites a quadratic expression from the form $a{x}^2+bx+c$ into the form $a(x+p{)}^2+q$. This is called the completed square form.

This is a Higher tier topic. On Foundation, algebraic quadratic solving is by factorising; approximate graphical solutions can also be assessed. Completing the square and the quadratic formula are Higher tier techniques.

The two main uses in GCSE Mathematics:

Use	How completing the square helps
Finding the turning point (vertex)	The turning point is directly readable from $(x+p{)}^2+q$
Solving quadratic equations	Rearrange the completed square form to isolate $x$

The method works because any quadratic with a squared term $x^2$ can be written as a squared bracket with a correction constant. The "completion" refers to turning a partial square into a perfect one.

Why learn it alongside the quadratic formula? Some exam questions specifically ask you to solve "by completing the square" or ask you to "write in the form $(x+p{)}^2+q$". The formula will not earn those marks.

When the coefficient of $x^2$ is 1, follow these three steps.

Step 1 — Halve the coefficient of $x$ and write it inside a squared bracket: $x^2+bx\to {\left(x+\frac{b}{2}\right)}^2$

Step 2 — Subtract the square of the number you just used (to correct the constant term that expanding introduces): ${\left(x+\frac{b}{2}\right)}^2-{\left(\frac{b}{2}\right)}^2$

Step 3 — Add on the original constant $c$: ${\left(x+\frac{b}{2}\right)}^2-{\left(\frac{b}{2}\right)}^2+c$

Worked example 1 — Complete the square for $x^2+6x+2$:

Half of 6 is 3, so the bracket is $(x+3{)}^2$.

$(x+3{)}^2-9+2=(x+3{)}^2-7$

Worked example 2 — Complete the square for $x^2-8x+3$:

Half of $-8$ is $-4$, so the bracket is $(x-4{)}^2$.

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

When the coefficient of $x^2$ is not 1, factor it out first before applying the standard method.

Step 1 — Factor out the coefficient of $x^2$ from the first two terms only.

Step 2 — Complete the square on the bracket inside.

Step 3 — Multiply back through and simplify.

Worked example — Complete the square for $2x^2+12x+5$:

Factor out 2 from the first two terms: $2(x^2+6x)+5$

Complete the square on $x^2+6x$. Half of 6 is 3: $2\left[(x+3{)}^2-9\right]+5$

Expand the outer factor and simplify: $2(x+3{)}^2-18+5=2(x+3{)}^2-13$

The correction constant you subtract inside the bracket must be multiplied by $a$ when you expand. Forgetting this multiplication is the most common error with $a\ne 1$ questions.

Once a quadratic is in completed square form $(x+p{)}^2+q$, the turning point (vertex) of its graph can be read off directly.

The minimum value of $(x+p{)}^2$ is 0, which occurs when $x=-p$. At that point, $y=q$.

Turning point: $(-p,q)$ from the form $(x+p{)}^2+q$.

Worked examples:

Completed square form	Turning point	Type
$(x+3{)}^2-7$	$(-3,-7)$	Minimum
$(x-4{)}^2-13$	$(4,-13)$	Minimum
$2(x+3{)}^2-13$	$(-3,-13)$	Minimum
$-(x-1{)}^2+5$	$(1,5)$	Maximum

When $a>0$ the parabola opens upward (minimum turning point). When $a<0$ it opens downward (maximum turning point). The sign of $a$ does not affect the $x$-coordinate of the vertex.

Worked example — Find the turning point of $y=x^2-8x+3$:

From the earlier worked example: $y=(x-4{)}^2-13$.

Turning point: $(4,-13)$, and it is a minimum (coefficient of $x^2$ is positive).

After completing the square, isolate the squared bracket and take the square root of both sides.

Worked example — Solve $x^2+6x+2=0$, giving your answers in surd form.

Complete the square: $(x+3{)}^2-7=0$

Rearrange: $(x+3{)}^2=7$

Take the square root (remember $\pm$): $x+3=\pm \sqrt{7}$

$x=-3+\sqrt{7}\approx -0.354\text{or}x=-3-\sqrt{7}\approx -5.646$

Worked example 2 — Solve $x^2-8x+3=0$, giving answers to 2 decimal places.

$(x-4{)}^2-13=0⟹(x-4{)}^2=13⟹x-4=\pm \sqrt{13}$

$x=4+\sqrt{13}\approx 7.61\text{or}x=4-\sqrt{13}\approx 0.39$

When taking the square root, always write $\pm$. Forgetting the negative root loses a mark and produces only one solution instead of two.

1. Halving incorrectly — halving the wrong term

The rule is: halve the coefficient of $x$, not the constant. In $x^2+10x+7$, the bracket is $(x+5{)}^2$, not $(x+10{)}^2$ or $(x+3.5{)}^2$.

2. Forgetting to subtract the squared constant

$(x+3{)}^2=x^2+6x+9$, not $x^2+6x$. The correction $-9$ must be subtracted to balance the equation. Omitting this step changes the expression's value.

3. Multiplying the correction by $a$ incorrectly

In $2(x+3{)}^2-9+5$, the $-9$ comes from subtracting $3^2=9$ inside the bracket, which is then multiplied by 2 to give $-18$. Writing $-9$ instead of $-18$ is the most common error in $a\ne 1$ problems.

4. Writing the turning point coordinates the wrong way round

From $(x+p{)}^2+q$, the $x$-coordinate of the turning point is $-p$ (note the sign change), and the $y$-coordinate is $q$. From $(x+3{)}^2-7$: turning point is $(-3,-7)$, not $(3,-7)$.

5. Forgetting $\pm$ when square-rooting

$\sqrt{7}$ has two values: $+\sqrt{7}$ and $-\sqrt{7}$. Writing only the positive root gives one solution instead of two, which costs a method mark and the second answer mark.