1.1 - Completing the square | Cambridge A-Level Mathematics

Two expressions can describe the same quadratic and reveal different things about it. In this first part of 1.1, you will rewrite a quadratic as a square plus a constant, then use that form to find its turning point and sketch its graph. The method also explains why some quadratics can never be negative.

Making a perfect square

Can $x^2+6x+2$ ever be less than $-7$ ? Substituting a few values might suggest an answer. Rewriting the expression will prove it for every real $x$ .

Start with a square you can recognise:

(x+3)^2=x^2+6x+9.

The first two terms match our expression. The constant is $7$ too large, so subtract $7$ :

x^2+6x+2=(x+3)^2-7.

The picture shows where the $9$ comes from. For a positive length $x$ , an $x$ -by- $x$ square and two $x$ -by- $3$ rectangles leave a $3$ -by- $3$ corner to fill. The algebraic identity remains true for negative $x$ too: expand the bracket to check it.

[DIAGRAM: asset_name: The missing corner; asset_slug: completing-square-area; caption: The missing corner has area 9, so x squared plus 6x is nine less than the whole square; description: A square of side x plus 3 is divided into an x by x square, two x by 3 rectangles and a 3 by 3 corner. The four areas are labelled x squared, 3x, 3x and 9, with side lengths x and 3 marked outside. The identity beneath is x squared plus 6x plus 9 equals the square of x plus 3.]
The missing corner has area 9, so x squared plus 6x is nine less than the whole square

Completing the square

Completing the square means rewriting a quadratic as a multiple of a squared linear expression plus a constant. The new expression has the same value as the original for every permitted input.

Why halve the coefficient of $x$ ? Because $(x+h)^2=x^2+2hx+h^2$ . To match $bx$ , choose $h=b/2$ , then correct the constant.

Completing a monic quadratic

Completing a monic quadratic.
For real $b$ and $c$ ,

x^2+bx+c=\left(x+\frac b2\right)^2+c-\frac{b^2}{4}.

Not in the formula booklet: learn it.

Use the reasoning behind the formula: match the linear term, then check the constant.

Worked example 1: a positive linear term

Question. Express $x^2+6x+2$ as a square plus a constant.

Plan. Half of $6$ is $3$ ; expand $(x+3)^2$ and adjust its constant.

\begin{aligned} x^2+6x+2&=(x+3)^2-9+2 &&\text{remove the added }9\\ &=(x+3)^2-7 &&\text{combine constants} \end{aligned}

Answer. $(x+3)^2-7$ .

Check. Expanding gives $x^2+6x+9-7=x^2+6x+2$ .

Now change just the sign of the linear term. The sign inside the bracket changes, but the correction still uses a positive square.

Worked example 2: a negative linear term

Question. Express $x^2-6x+2$ as a square plus a constant.

Plan. Half of $-6$ is $-3$ , so start with $(x-3)^2$ .

\begin{aligned} x^2-6x+2&=(x-3)^2-9+2 &&\text{correct the constant}\\ &=(x-3)^2-7 \end{aligned}

Answer. $(x-3)^2-7$ .

Check. At $x=3$ , the original gives $9-18+2=-7$ , and the new form gives $0-7=-7$ . Expanding confirms equality for every $x$ .

The same method works with odd coefficients: for $x^2+5x$ , the bracket starts $x+5/2$ . Keep the fraction exact.

Watch out

$(x-3)^2$ is $x^2-6x+9$ , not $x^2-9$ . A squared bracket has a middle term. Expanding is a quick way to catch a wrong sign or constant.

Try the same method with a different linear coefficient.

A coefficient before the square

When the coefficient of $x^2$ is not $1$ , factor it out of the first two terms before completing the square. Anything you add or subtract inside that bracket is multiplied by the outside coefficient.

For example, $2(x^2+6x)$ becomes $2[(x+3)^2-9]$ . The correction outside the square is $-18$ , not $-9$ .

The general completed-square form

The general completed-square form.
For $a\ne0$ ,

ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.

Not in the formula booklet: learn it.

You can reconstruct this result by factoring and expanding; there is no need to force every calculation through a memorised substitution.

Worked example 3: a positive multiplier

Question. Express $2x^2+12x+7$ in the form $2(x+p)^2+q$ .

Plan. Factor $2$ from the first two terms, complete the square inside, then multiply out the correction.

\begin{aligned} 2x^2+12x+7&=2(x^2+6x)+7\\ &=2[(x+3)^2-9]+7\\ &=2(x+3)^2-11 \end{aligned}

Answer. $2(x+3)^2-11$ , so $p=3$ and $q=-11$ .

Check. Expanding gives $2x^2+12x+18-11$ , which is the original expression.

Keep $x^2+6x$ and the constant $7$ , but change the multiplier from $2$ to $-2$ . This changes the sign of the correction outside the bracket.

Worked example 4: a negative multiplier

Question. Express $-2x^2-12x+7$ in the form $-2(x+p)^2+q$ .

Plan. Factor $-2$ , then complete the same square as before.

\begin{aligned} -2x^2-12x+7&=-2(x^2+6x)+7\\ &=-2[(x+3)^2-9]+7\\ &=-2(x+3)^2+25 \end{aligned}

Answer. $-2(x+3)^2+25$ , so $p=3$ and $q=25$ .

Check. At $x=0$ , the new form gives $-18+25=7$ , matching the original constant. Expansion also recovers $-12x$ .

The outside coefficient acts on the whole bracket, including the correction.

Watch out

$2[(x+3)^2-9]\ne2(x+3)^2-9$ . The correct expression is $2(x+3)^2-18$ . Leave the square bracket in place until you have multiplied both terms.

Here is another negative multiplier to handle.

The turning point

A real square is non-negative, and it equals zero exactly when its base is zero. That simple fact reveals the highest or lowest point of a quadratic graph.

In $y=2(x-3)^2-5$ , the squared term is smallest at $x=3$ . Its value there is zero, so the smallest possible $y$ is $-5$ . Inputs equally far from $3$ , such as $2$ and $4$ , give the same squared term and the same height.

Vertex and axis of symmetry

The vertex of a quadratic graph is its turning point. For $y=a(x-h)^2+k$ , where $a\ne0$ , the vertex is $(h,k)$ and the axis of symmetry is the vertical line $x=h$ .

If $a>0$ , the graph opens upwards and $k$ is its minimum value. If $a<0$ , multiplying a square by $a$ makes it non-positive: the graph opens downwards and $k$ is its maximum value. These statements allow every real input $x$ .

[DIAGRAM: asset_name: One vertex, two directions; asset_slug: quadratic-vertex-sign; caption: The sign before the square decides whether the vertex is a minimum or a maximum; description: Two coordinate plots with matching scales show y equals 2 times the square of x minus 3 minus 5 and y equals minus 2 times the square of x minus 3 minus 5. Both have vertex 3 comma minus 5 and symmetry line x equals 3. The first opens upwards and labels minimum minus 5; the second opens downwards and labels maximum minus 5.]
The sign before the square decides whether the vertex is a minimum or a maximum

Worked example 5: the minimum value

Question. State the minimum value of $2(x-3)^2-5$ and the value of $x$ at which it occurs.

Plan. Use the fact that the square is at least zero and identify when it vanishes.

2(x-3)^2\ge0 \quad\Longrightarrow\quad 2(x-3)^2-5\ge-5.

Equality holds when $x-3=0$ .

Answer. Minimum value $-5$ , at $x=3$ . The vertex is $(3,-5)$ .

Check. At the neighbouring inputs $x=2$ and $x=4$ , the value is $-3$ , higher than $-5$ .

Now change only the sign of the coefficient before the square. The vertex stays where it was, but its role changes.

Worked example 6: the maximum value

Question. State the maximum value of $-2(x-3)^2-5$ and the value of $x$ at which it occurs.

Plan. A negative multiple of a square cannot exceed zero.

-2(x-3)^2\le0 \quad\Longrightarrow\quad -2(x-3)^2-5\le-5.

Equality again holds when $x=3$ .

Answer. Maximum value $-5$ , at $x=3$ . The vertex is $(3,-5)$ .

Check. At $x=2$ and $x=4$ , the value is $-7$ , below the maximum.

Read the horizontal coordinate by asking which input makes the bracket zero.

Watch out

For $y=(x+4)^2+9$ , the vertex is $(-4,9)$ , not $(4,9)$ . The bracket vanishes at $x=-4$ .

Use both the bracket and its multiplier in this check.

Sketching from the square

A completed square gives you the shape and turning point immediately. Add the $y$ -intercept by substituting $x=0$ , then use symmetry to place a second point. This gives a useful sketch without a long table of values.

Worked example 7: a graph above the axis

Question. Find the vertex and $y$ -intercept of $y=x^2-4x+7$ , and describe its position relative to the $x$ -axis.

Plan. Complete the square, then use its minimum and substitute $x=0$ .

y=(x-2)^2-4+7=(x-2)^2+3.

The graph opens upwards. Its minimum is $3$ , attained at $x=2$ , and its symmetry line is $x=2$ . At $x=0$ , $y=7$ ; symmetry gives another point $(4,7)$ .

Answer. Vertex $(2,3)$ , $y$ -intercept $(0,7)$ . The whole graph lies above the $x$ -axis because $y\ge3$ .

Check. Direct substitution gives $y(2)=4-8+7=3$ and $y(4)=16-16+7=7$ .

The three labelled points and the symmetry line locate the curve clearly.

[DIAGRAM: asset_name: A sketch from three features; asset_slug: quadratic-sketch; caption: A completed square locates the vertex and explains why this graph stays above the x-axis; description: Coordinate plot of y equals x squared minus 4x plus 7, with vertex 2 comma 3, y-intercept 0 comma 7, reflected point 4 comma 7 and dashed symmetry line x equals 2. The curve opens upwards and its minimum height is 3.]
A completed square locates the vertex and explains why this graph stays above the x-axis

Next, keep the first two terms but make the constant unknown. An extreme value can tell us what that constant must be.

Worked example 8: an unknown constant

Question. The minimum value of $x^2-4x+k$ , for real $x$ , is $8$ . Find $k$ and state the vertex of its graph.

Plan. Complete the square in terms of $k$ , then equate the minimum to $8$ .

x^2-4x+k=(x-2)^2+k-4.

The square is zero at $x=2$ , so the minimum is $k-4$ . Therefore

k-4=8\quad\Longrightarrow\quad k=12.

Answer. $k=12$ , with vertex $(2,8)$ .

Check. Substituting $x=2$ into $x^2-4x+12$ gives $8$ . Every other real input adds a positive square to $8$ .

Return to our opening question: $x^2+6x+2=(x+3)^2-7$ , so it cannot be less than $-7$ . It reaches $-7$ exactly at $x=-3$ . One rearrangement has settled infinitely many possible inputs.

Exam tip

When asked for a particular completed-square form, give that form explicitly. Expand it once to check the constant correction: this is the check highlighted in Cambridge's guidance on this question type.

Use the minimum to recover a missing constant in a new quadratic.

Quick Recap

Match a square by halving the linear coefficient, then correct the constant.
Factor the coefficient of $x^2$ first when it is not $1$ .
In $y=a(x-h)^2+k$ , the vertex is $(h,k)$ . It is a minimum for $a>0$ and a maximum for $a<0$ .
Use the vertex, symmetry and $y$ -intercept to sketch a quadratic, and expand your completed square to check it.

The opening quadratic cannot fall below $-7$ : $x^2+6x+2=(x+3)^2-7$ .

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1.1 - Completing the square

Making a perfect square

A coefficient before the square

The turning point

Sketching from the square

Quick Recap

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