B4PureCore

Completing the Square

Q: Why does completing the square give the turning point?

In the form a(x+p)^2 + q, the squared bracket is smallest (zero) when x = -p, so the curve turns there, at height q. That makes (-p, q) the vertex and q the minimum (or maximum if a<0).

Q: How do I handle a number in front of x^2?

Factor it out of the x^2 and x terms first: 2x^2 + 8x + 3 = 2(x^2+4x)+3. Complete the square inside the bracket, expand back, and combine constants. Leave the lone constant outside until the end.

Q: How does this prove a quadratic is always positive?

If the completed form is a(x+p)^2 + q with a>0 and q>0, then since (x+p)^2 0 the whole thing is at least q>0, so it can never be zero or negative — exactly the deduction-proof technique.

Completing the square rewrites a quadratic as a perfect square plus a constant, $a (x + p)^{2} + q$ . That single rearrangement hands you the turning point, the minimum value, exact surd solutions and a way to prove a quadratic is always positive.

30 min Video by TLMaths Algebra and Functions

The big picture

Factorising and the formula both find the roots — but completing the square tells you the $shape$ : where the parabola turns, how low it dips, and whether it ever touches the axis at all. It is the most information-rich thing you can do to a quadratic, which is why it powers so much else: the quadratic formula is literally completing the square done in general; circle equations are completed squares in $x$ and $y$ ; and the “prove it is always positive” arguments from the Proof chapter are completing the square in disguise. Learn to do it fluently and a whole web of later topics becomes routine.

A-Level Maths B3-08: Introducing Completing the SquareWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Complete the square for $x^{2} + b x + c$ and for $a x^{2} + b x + c$
Read the turning point and minimum/maximum straight from the completed form
Solve a quadratic in exact surd form by completing the square
Use the completed square to prove a quadratic is always positive

The idea: a quadratic is almost a perfect square

Notice that $(x + p)^{2} = x^{2} + 2 p x + p^{2}$ . So a perfect square always has a constant term equal to the $\textbf{square of half the$ x $-coefficient}$ . A general quadratic $x^{2} + b x + c$ usually has the “wrong” constant — so we build the perfect square we want, then add a correction to fix the constant.

Concretely, take half of $b$ , square it, and adjust: $x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}$ . The subtracted term undoes the extra constant the square introduced.

x^{2} + b x + c = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c

Halve

b

, square it, subtract it back.

Why this works

Everything hinges on “half the coefficient of $x$ ”. The square $(x + \frac{b}{2})^{2}$ is engineered so its middle term is $b x$ ; the $- (\frac{b}{2})^{2}$ simply cancels the constant that the square smuggled in.

The method in action

Work in three moves: write the square using half of $b$ , subtract the square of that half, then tidy the constants.

1Half of 6 is 3, so start with

(x + 3)^{2}

2That expands to

x^{2} + 6 x + 9

— an extra 9 — so subtract it:

(x + 3)^{2} - 9

3Add the original constant:

(x + 3)^{2} - 9 + 1 = (x + 3)^{2} - 8

Answer

(x + 3)^{2} - 8

Tip — The sign inside the bracket copies the sign of $b$ : $x^{2} - 5 x$ gives $(x - \frac{5}{2})^{2}$ , not $(x + \frac{5}{2})^{2}$ .

When the $x^{2}$ coefficient isn’t 1

If $a \neq = 1$ , factor $a$ out of the $x^{2}$ and $x$ terms first, complete the square inside the bracket, then multiply back out. Keep the constant outside the bracket until the end.

1Factor 2 from the first two terms:

2 (x^{2} + 4 x) + 3

2Inside: half of 4 is 2, so

x^{2} + 4 x = (x + 2)^{2} - 4

3So

2 [(x + 2)^{2} - 4] + 3 = 2 (x + 2)^{2} - 8 + 3

4Tidy:

2 (x + 2)^{2} - 5

Answer

2 (x + 2)^{2} - 5

Tip — Only factor $a$ from the $x^{2}$ and $x$ terms — leave the lone constant alone until the very end.

What the completed form tells you

Once a quadratic is $a (x + p)^{2} + q$ , its secrets are on display. Because a square is never negative, $(x + p)^{2} \geq 0$ , so the whole expression is smallest when the bracket is zero — at $x = - p$ . That point $(- p, q)$ is the $turning point$ (vertex), and $q$ is the minimum value (or maximum if $a < 0$ ).

This is also how you solve in exact form (rearrange for the bracket, square-root both sides — remembering $\pm$ ) and how you prove positivity: if $q > 0$ and $a > 0$ , the quadratic can never reach zero, so it is always positive — exactly the argument used in Proof by Deduction.

a (x + p)^{2} + q \Rightarrow turning point (- p, q)

Read the vertex straight off — no calculus needed.

1From before,

x^{2} + 6 x + 1 = (x + 3)^{2} - 8

, so

(x + 3)^{2} - 8 = 0

2Rearrange:

(x + 3)^{2} = 8

3Square-root both sides (with

\pm

x + 3 = \pm 8 = \pm 22

4So

x = - 3 \pm 22

Answer

x = - 3 \pm 22

Why this works

The completed square finds the turning point without differentiating. Later, calculus gives the same vertex by setting the derivative to zero — it is reassuring, and no accident, that two very different methods agree.

Think like an examiner

A completed-square answer $a (x + p)^{2} + q$ is often followed by “hence write down the turning point” or “the minimum value” — you should be able to read them off with no extra work.
“Prove $\dots > 0$ for all $x$ ” is a completing-the-square question in disguise: show it equals a square plus a positive constant.
When $a \neq = 1$ , marks are lost by not factoring $a$ out first, or by mishandling the constant. Keep it outside the bracket.
For exact solutions, do not forget the $\pm$ when you square-root, and simplify the surd (e.g. $8 = 22$ ).

Common misconceptions

Forgetting to subtract

(\frac{b}{2})^{2}

after forming the square.

The square adds an extra

(\frac{b}{2})^{2}

; you must subtract it to keep the expression equal.

Completing the square on

2 x^{2} + 8 x + 3

without factoring the 2 first.

Factor:

2 (x^{2} + 4 x) + 3

, complete inside, then expand back — otherwise the

x^{2}

coefficient is wrong.

Writing

(x - 4)^{2}

for

x^{2} + 4 x

(copying the sign of

b

wrongly).

Half of

+ 4

+ 2

, so it is

(x + 2)^{2} - 4

Solving

(x + 3)^{2} = 8

x + 3 = 8

only.

Square-rooting gives

\pm

x + 3 = \pm 22

, so there are two solutions.

Completing the square

x^{2} + b x + c = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c

The core rearrangement.

a (x + p)^{2} + q

Vertex form: turning point

(- p, q)

(x + p)^{2} \geq 0

Why

q

is the minimum.

Stretch yourself

Complete the square on the general quadratic $a x^{2} + b x + c = 0$ to derive the quadratic formula.

Hint — Divide through by $a$ , complete the square on $x^{2} + \frac{b}{a} x$ , then rearrange and square-root (with $\pm$ ).

Questions students ask

Why does completing the square give the turning point?

In the form

a (x + p)^{2} + q

, the squared bracket is smallest (zero) when

x = - p

, so the curve turns there, at height

q

. That makes

(- p, q)

the vertex and

q

the minimum (or maximum if

a < 0

How do I handle a number in front of

x^{2}

Factor it out of the

x^{2}

and

x

terms first:

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

. Complete the square inside the bracket, expand back, and combine constants. Leave the lone constant outside until the end.

When should I complete the square instead of using the formula?

When you need the turning point, the minimum/maximum, an exact surd solution, or to prove a quadratic is always positive. For roots alone, factorising or the formula is usually quicker.

How does this prove a quadratic is always positive?

If the completed form is

a (x + p)^{2} + q

with

a > 0

and

q > 0

, then since

(x + p)^{2} \geq 0

the whole thing is at least

q > 0

, so it can never be zero or negative — exactly the deduction-proof technique.

Key takeaways

Complete the square by halving $b$ , squaring it and subtracting it back.
If $a \neq = 1$ , factor $a$ from the $x^{2}$ and $x$ terms first.
The form $a (x + p)^{2} + q$ gives the turning point $(- p, q)$ and the min/max value $q$ .
Solve in exact form by rearranging and square-rooting (keep the $\pm$ ).
A completed square with $a > 0$ , $q > 0$ proves the quadratic is always positive.

How this fits the course

Build on

The Discriminant
Circles (later)soon

Leads to

The Discriminant

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