2A - Completing the square

How small can a quadratic expression become? You will rewrite quadratics as a multiple of one square plus a constant, making their minimum or maximum visible without plotting lots of points.

Making a perfect square

How small can $x^2+6x+5$ become as $x$ varies over all real numbers?

At $x=0$ it is $5$ ; at $x=-3$ it is $-4$ ; at $x=-6$ it is $5$ again. Those values suggest a lowest point near $-3$ , but checking a few values cannot prove a minimum. We need a form that tells us about every real $x$ .

Expand one square:

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

It has the right first two terms but the constant is $4$ too large. Subtracting $4$ gives

x^2+6x+5=(x+3)^2-4.

A square cannot be negative, so this expression is at least $-4$ , with equality when $x=-3$ .

Completing the square

Completing the square means rewriting a quadratic as a single squared linear expression, possibly multiplied by a constant, plus a remaining constant. The new expression must be equal to the original for every value of the variable.

The coefficient of $x$ in $(x+p)^2=x^2+2px+p^2$ is twice $p$ . That is why we halve the middle coefficient.

Completing a monic square

For real constants $b,c$ ,

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

Halve $b$ , square that half, then compensate for the constant introduced.

Not on the formulae sheet: learn it.

Use the identity as a reason for the steps, rather than as a long formula to substitute into mechanically.

Worked example 1: an even middle coefficient

Question. Express $x^2+6x+5$ in completed-square form.

Plan. Use $(x+3)^2$ , then correct its constant term.

\begin{aligned} x^2+6x+5&=(x+3)^2-9+5 &&\text{add and subtract }3^2\\ &=(x+3)^2-4 &&\text{collect the constants} \end{aligned}

Answer. $(x+3)^2-4$ .

Check. Expansion gives $x^2+6x+9-4=x^2+6x+5$ .

Change only the middle coefficient from $6$ to $5$ . Its half is a fraction; the method is unchanged.

Worked example 2: an odd middle coefficient

Question. Express $x^2+5x+5$ in the form $(x+p)^2+q$ .

Plan. Halve $5$ and subtract the square of that half before adding the original constant.

\begin{aligned} x^2+5x+5&=\left(x+\frac52\right)^2-\frac{25}{4}+5 &&\text{compensate for }\left(\frac52\right)^2\\ &=\left(x+\frac52\right)^2-\frac54 &&\text{write }5=\frac{20}{4} \end{aligned}

Answer. $p=\frac52$ , $q=-\frac54$ , giving $\left(x+\frac52\right)^2-\frac54$ .

Check. The constant after expansion is $\frac{25}{4}-\frac54=5$ , and the middle term is $2\times\frac52x=5x$ .

The square of a bracket contains a cross term.

Watch out

$(x+3)^2\ne x^2+9$ : the expansion is $x^2+6x+9$ . Also subtract the square of the half-coefficient, not the half-coefficient itself.

Keep fractions exact in the next expression.

Other leading coefficients

A leading coefficient other than one must be taken out before you complete the square inside a bracket.

For $2x^2+6x+1$ , first write $2(x^2+3x)+1$ . The middle coefficient inside is now $3$ , not $6$ . Every constant introduced inside that bracket is multiplied by the outside $2$ .

For any $a\ne0$ , the first step is

ax^2+bx+c=a\left(x^2+\frac ba x\right)+c.

You may instead take $a$ out of all three terms, provided you divide the constant by $a$ too. Leaving the constant outside often keeps the arithmetic simpler.

Worked example 3: a positive leading coefficient

Question. Express $2x^2+6x+1$ in the form $a(x+p)^2+q$ .

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

\begin{aligned} 2x^2+6x+1&=2(x^2+3x)+1 &&\text{factor the leading coefficient}\\ &=2\left[\left(x+\frac32\right)^2-\frac94\right]+1 &&\text{halve }3\text{ inside}\\ &=2\left(x+\frac32\right)^2-\frac92+1 &&\text{multiply the correction by }2\\ &=2\left(x+\frac32\right)^2-\frac72 &&\text{collect constants} \end{aligned}

Answer. $2\left(x+\frac32\right)^2-\frac72$ .

Check. Expansion gives $2x^2+6x+\frac92-\frac72=2x^2+6x+1$ .

Keep the middle and constant terms but change the leading coefficient from $2$ to $-\frac12$ . This tests both a negative outside factor and division by a fraction.

Worked example 4: a negative fractional coefficient

Question. Express $-\frac12x^2+6x+1$ in completed-square form.

Plan. Take out $-\frac12$ from the first two terms, so the bracket starts with $x^2$ .

\begin{aligned} -\frac12x^2+6x+1&=-\frac12(x^2-12x)+1 &&\text{since }6\div\left(-\frac12\right)=-12\\ &=-\frac12\big[(x-6)^2-36\big]+1 &&\text{complete the square inside}\\ &=-\frac12(x-6)^2+18+1 &&\text{multiply both bracket terms}\\ &=-\frac12(x-6)^2+19 &&\text{collect constants} \end{aligned}

Answer. $-\frac12(x-6)^2+19$ .

Check. Expanding gives $-\frac12x^2+6x-18+19=-\frac12x^2+6x+1$ . In particular, $x=0$ gives $1$ in both forms.

The outside factor is part of the expression, so it affects the correction as well as the square.

Watch out

$2[(x+\frac32)^2-\frac94]+1$ becomes $2(x+\frac32)^2-\frac92+1$ , not $2(x+\frac32)^2-\frac94+1$ . Distribute the outside coefficient to every term.

Apply that same distribution carefully with a negative fraction.

Reading the completed-square form

A completed square locates a quadratic graph's turning point because it shows where the squared quantity is zero.

In $y=(x+3)^2-4$ , putting $x=-3$ makes the square zero and gives $y=-4$ . Moving the same distance to either side gives the same $y$ : for instance, $x=-2$ and $x=-4$ both give $-3$ . The graph is symmetric about $x=-3$ .

Vertex

The vertex of a quadratic graph is its turning point: its lowest point when the leading coefficient is positive, or its highest point when that coefficient is negative. The vertical line through it is the axis of symmetry.

For $y=a(x-h)^2+k$ , where $a\ne0$ , the square is zero at $x=h$ . The vertex is therefore $(h,k)$ and the symmetry axis is $x=h$ . If $a>0$ , multiplying a nonnegative square by $a$ keeps it nonnegative, so $y\ge k$ . If $a<0$ , the inequality reverses and $y\le k$ . In either case the bound is attained at the vertex.

[DIAGRAM: asset_name: One vertex with a minimum or a maximum; asset_slug: completed-square-vertices; caption: Changing the sign of the squared term changes a minimum into a maximum at the same vertex; description: Two coordinate plots with the same scales compare y equals 3 times x minus 2 squared minus 5 and y equals minus 3 times x minus 2 squared minus 5. Both have vertex 2 comma minus 5 and symmetry line x equals 2. The first opens upward and has minimum minus 5; the second opens downward and has maximum minus 5. Equal-distance points on either side of x equals 2 are shown at x equals 1 and x equals 3.]
Changing the sign of the squared term changes a minimum into a maximum at the same vertex

The completed-square form also describes graph transformations: shift the graph of $y=ax^2$ horizontally by $h$ and vertically by $k$ . A negative $a$ reflects its opening downwards.

Worked example 5: a minimum

Question. For $y=3(x-2)^2-5$ , find the vertex, axis of symmetry and minimum value.

Plan. Make the square zero and use the fact that a positive multiple of a square is nonnegative.

\begin{aligned} (x-2)^2&\ge0 &&\text{every real square is nonnegative}\\ 3(x-2)^2-5&\ge-5 &&\text{multiply by }3\text{ and subtract }5\\ x=2&\implies y=-5 &&\text{the bound is attained} \end{aligned}

Answer. Vertex $(2,-5)$ , axis $x=2$ , minimum value $-5$ .

Check. At $x=1$ and $x=3$ , the value is $-2$ , equally above the vertex. Expanding gives $3x^2-12x+7$ , so the leading coefficient is positive as expected.

Change only the multiplier of the square from $3$ to $-3$ . The location where the square is zero stays fixed, but the bound becomes a maximum.

Worked example 6: a maximum

Question. For $y=-3(x-2)^2-5$ , find the vertex, axis of symmetry and maximum value.

Plan. Multiply the nonnegative square by the negative coefficient before interpreting the bound.

\begin{aligned} (x-2)^2&\ge0 &&\text{start with the square}\\ -3(x-2)^2&\le0 &&\text{multiplication by a negative reverses the sign}\\ y=-3(x-2)^2-5&\le-5 &&\text{subtract }5 \end{aligned}

Equality occurs at $x=2$ .

Answer. Vertex $(2,-5)$ , axis $x=2$ , maximum value $-5$ .

Check. At $x=1$ and $x=3$ , $y=-8$ , equally below the vertex. Expansion gives $-3x^2+12x-17$ , with negative leading coefficient.

Read the sign inside the bracket by asking what value makes it zero.

Watch out

For $y=(x+3)^2-4$ , the vertex is $(-3,-4)$ , not $(3,-4)$ . The minimum value is the $y$ -coordinate $-4$ ; the value of $x$ where it occurs is $-3$ .

This proves the opening expression has minimum $-4$ , rather than merely suggesting it from a few plotted values.

Quick Recap

Halve the middle coefficient inside a monic quadratic, then compensate for its square.
For another leading coefficient, factor it out first and multiply the constant correction by it.
The form $a(x-h)^2+k$ has vertex $(h,k)$ and symmetry axis $x=h$ .
Positive $a$ gives minimum $k$ ; negative $a$ gives maximum $k$ , for all real $x$ .

The opening expression is $(x+3)^2-4$ , so its minimum is exactly $-4$ at $x=-3$ .

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2A - Completing the square

Making a perfect square

Other leading coefficients

Reading the completed-square form

Quick Recap

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