What you'll learn
- How to turn expressions with an x2x^2x2 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 x2x^2x2.
- How to solve equations exactly using square roots.
Key vocabulary
Essential terms
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Completing the square means rewriting an expression like x2+px+qx^2+px+qx2+px+q so it contains a squared bracket.
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A quadratic expression has highest power x2x^2x2; its graph is a parabola, a U-shaped curve.
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A coefficient is the number multiplying a variable; a constant is a number on its own.
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A perfect square is a squared bracket such as (x−3)2(x-3)^2(x−3)2.
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Completed square form is (x+h)2+k(x+h)^2+k(x+h)2+k or a(x+h)2+ka(x+h)^2+ka(x+h)2+k, where the letters stand for constants.
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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=x2+2rx+r2(x+r)^2=x^2+2rx+r^2(x+r)2=x2+2rx+r2So completing the square is really running this expansion backwards. The middle term, 2rx2rx2rx, tells you that the number in the bracket is half the coefficient of xxx.

Half, then square
For x2+px+qx^2+px+qx2+px+q, half the coefficient of xxx to make the bracket, then subtract the square you accidentally added.

Reversing the expansion
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The coefficient of xxx in x2+10xx^2+10xx2+10x is 10, so half is 5.
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Build the square bracket and expand it:
(x+5)2=x2+10x+25(x+5)^2=x^2+10x+25(x+5)2=x2+10x+25 -
The square contains 25 too much, so subtract 25:
x2+10x=(x+5)2−25x^2+10x=(x+5)^2-25x2+10x=(x+5)2−25
2. Completing the square when the x2x^2x2 coefficient is 1
For an expression in the form x2+px+qx^2+px+qx2+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)^2x2+px+q=(x+2p)2+q−(2p)2You do not need to memorise this as a “big formula” if you remember the half-and-square idea.
Rewrite and find the turning point
Rewrite x2−8x+9x^2-8x+9x2−8x+9 in completed square form, then find the turning point of y=x2−8x+9y=x^2-8x+9y=x2−8x+9.

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The coefficient of xxx is -8, and half of -8 is -4.
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Create the bracket (x−4)2(x-4)^2(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}x2−8x+9=(x−4)2−16+9=(x−4)2−7 -
So the completed square form is (x−4)2−7(x-4)^2-7(x−4)2−7.
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The square part is smallest when x=4x=4x=4, so the turning point is (4,−7)(4,-7)(4,−7).
Subtracting the half, not the square
If the bracket is (x−4)2(x-4)^2(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+ky=(x+h)^2+ky=(x+h)2+k, the square is always zero or positive. So the lowest value happens when the bracket equals zero.
That means:
y=(x+h)2+khas turning point(−h,k)y=(x+h)^2+k \quad \text{has turning point} \quad (-h,k)y=(x+h)2+khas turning point(−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.
Opposite sign in the bracket
The x-coordinate uses the opposite sign to the bracket number: (x+6)2(x+6)^2(x+6)2 gives x=−6x=-6x=−6.

4. Odd coefficients often give fractions
If the coefficient of xxx is odd, halving it gives a fraction. That is completely normal at higher GCSE.
Completing the square with fractions
Find the turning point of y=x2−3x+2y=x^2-3x+2y=x2−3x+2.

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Half of -3 is −32-\frac{3}{2}−23, so the bracket is (x−32)2\left(x-\frac{3}{2}\right)^2(x−23)2.
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Square the half-coefficient:
(−32)2=94\left(-\frac{3}{2}\right)^2=\frac{9}{4}(−23)2=49 -
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}x2−3x+2=(x−23)2−49+2=(x−23)2−41 -
The turning point is (32,−14)\left(\frac{3}{2},-\frac{1}{4}\right)(23,−41).
5. When the coefficient of x2x^2x2 is not 1
The leading coefficient is the coefficient of x2x^2x2. If it is not 1, factor it out of the x2x^2x2 and xxx terms first, then complete the square inside the bracket.
A coefficient in front of
Rewrite 2x2+16x+352x^2+16x+352x2+16x+35 in the form a(x+b)2+ca(x+b)^2+ca(x+b)2+c, then state the turning point.

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Factor 2 from the first two terms:
2x2+16x+35=2(x2+8x)+352x^2+16x+35=2(x^2+8x)+352x2+16x+35=2(x2+8x)+35 -
Complete the square inside the bracket:
x2+8x=(x+4)2−16x^2+8x=(x+4)^2-16x2+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}2(x2+8x)+35=2((x+4)2−16)+35=2(x+4)2−32+35=2(x+4)2+3 -
The turning point is (−4,3)(-4,3)(−4,3).
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.
Surd form
A surd is an exact root that is not a whole number, such as 6\sqrt{6}6; surd form means leaving it exact instead of rounding.
Solving exactly with a decimal constant
Solve x2+7x+6.25=0x^2+7x+6.25=0x2+7x+6.25=0 by completing the square.

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Write 6.25 as an exact fraction:
6.25=2546.25=\frac{25}{4}6.25=425 -
Half of 7 is 72\frac{7}{2}27, 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}x2+7x+425=(x+27)2−449+425=(x+27)2−6 -
Set the completed square expression equal to zero:
(x+72)2−6=0\left(x+\frac{7}{2}\right)^2-6=0(x+27)2−6=0 -
Isolate the square:
(x+72)2=6\left(x+\frac{7}{2}\right)^2=6(x+27)2=6 -
Square root both sides using ±\pm±, then subtract 72\frac{7}{2}27: x=−72±6x=-\frac{7}{2}\pm\sqrt{6}x=−27±6.
Missing one solution
When you square root both sides, remember the ±\pm± sign. Without it, you only find one of the two roots.
In the exam
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Show the halving step clearly, because it earns method marks.
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For turning points, read the x-coordinate with the opposite sign to the bracket.
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When solving, isolate the squared bracket before square rooting, and leave exact roots in surd form.
Check yourself
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Complete the square for x2+12x−5x^2+12x-5x2+12x−5.
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What is the turning point of y=2(x−3)2+7y=2(x-3)^2+7y=2(x−3)2+7?
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Why must you write ±\pm± when solving (x+1)2=11(x+1)^2=11(x+1)2=11?