Completing the square: worked examples and method

Higher · Grade 8 · Non-calculator

This page sets out the method for completing the square in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Completing the square topic page.

Key facts

$x^{2} + b x + c = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c$ .
The graph of $y = (x + p)^{2} + q$ has its turning point at $(- p, q)$ — the sign of $p$ flips, the sign of $q$ does not.
If the $x^{2}$ coefficient is positive the turning point is a minimum and $q$ is the minimum value.
For $a x^{2} + b x + c$ with $a \neq = 1$ , write $a (x^{2} + \frac{b}{a} x) + c$ first, then complete the square inside.
From $(x + p)^{2} = k$ : $x = - p \pm k$ . Keep the surd exact unless the question says otherwise.
The quadratic formula is completing the square done in general — both give the same exact answers.

The method, step by step

Method: Coefficient of $x^{2}$ is 1

When you see 'write $x^{2} + b x + c$ in the form $(x + a)^{2} + b$ ', do this.

Halve the coefficient of $x$ — this is $p$ .
Write $(x + p)^{2}$ .
Subtract $p^{2}$ (because expanding the bracket added it in).
Add the original constant $c$ and simplify the numbers.
Check by expanding your answer back out.

Method: Coefficient of $x^{2}$ is not 1

When you see $2 x^{2} - 12 x + 7$ or $3 x^{2} + 6 x - 1$ , do this.

Factorise the coefficient $a$ out of the $x^{2}$ and $x$ terms only, using a square bracket.
Complete the square inside the bracket as normal.
Multiply the subtracted square by $a$ when you take it outside the bracket.
Combine with the original constant.
Write the answer as $a (x + p)^{2} + q$ .

Method: Solve exactly using the completed square

When you see 'solve … giving your answer in the form $a \pm b$ ' or 'in surd form', do this.

Complete the square on the left-hand side with the equation $= 0$ .
Move the constant to the right so the squared bracket is alone.
Square-root both sides and write $\pm$ .
Subtract $p$ from both sides.
Simplify the surd if possible (e.g. $12 = 23$ ).

How the lesson explains it

Half, then square

For x^2 + bx, the bracket uses b/2. You then subtract (b/2)^2 so the expression is unchanged.

Read the turning point

(x+p)^2+q has its turning point at (-p, q). The sign inside the bracket flips.

Take out a

If the x^2 term is 2x^2, factor 2 from the x terms first. The constant outside the factor stays separate.

a is not always 1

2(x+2)^2-3 is not the same as (2x+2)^2-3. The 2 multiplies the whole square.

Solve from the square

Once you have a(x+p)^2+q=0, divide by a, then square-root both sides. You get two roots.

Minimum or maximum

If a is positive the turning value is a minimum. If a is negative it is a maximum.

Halve, square, adjust

For x^2+6x+2, half of 6 is 3, and 3^2=9. Write (x+3)^2-9+2=(x+3)^2-7. The -9 cancels the extra 9 that the square introduced, and the +2 is still there. At x=1 the original is 1+6+2=9 and the new form is 16-7=9. The turning point is (-3, -7), because the square is zero at x=-3 and the value is -7. Check: (-3)^2+6(-3)+2=9-18+2=-7.

A negative b puts a minus inside the bracket. x^2-8x+5=(x-4)^2-16+5=(x-4)^2-11. Half of -8 is -4, and the square of -4 is still 16, which you subtract. The turning point is (4, -11), not (-4, -11). At x=2 both forms give 4-16+5=-7 and 4-11=-7. If those disagree, the constant adjustment is wrong.

Method: 1. Halve the coefficient of x. 2. Square that half. 3. Add and subtract it so the expression is unchanged. 4. Write (x+p)^2 minus the adjustment, including the original constant.

When a is not 1

For 2x^2+8x-5, factor 2 out of the first two terms only: 2(x^2+4x)-5. Complete the square inside: x^2+4x=(x+2)^2-4. Then 2((x+2)^2-4)-5=2(x+2)^2-8-5=2(x+2)^2-13. The -4 was multiplied by 2 to become -8. Forgetting that multiplication is the usual error when a≠ 1. The turning point is (-2, -13).

Check at x=1: the original is 2+8-5=5, and 2(3)^2-13=18-13=5. The minimum value is -13, at x=-2. Because the coefficient of x^2 is positive, the parabola opens upwards and -13 really is the lowest value. You can say "2x^2+8x-5 is always at least -13" and the completed square is the reason.

Method: 1. Factor the coefficient of x^2 out of the x^2 and x terms only. 2. Complete the square on the bracket that is left. 3. Multiply the adjustment back by the coefficient you took out. 4. Check by expanding, or by substituting one value of x.

Solve by isolating the square

From (x+3)^2-7=0, add 7 and square root: x+3=±√(7), so x=-3±√(7). Both signs are required. The same equation is x^2+6x+2=0, whose discriminant is 36-8=28=4× 7, and the formula gives x=dfrac-6± 2√(7)2=-3±√(7). The two methods agree. Use completing the square when the question asks for it, or when you also need the turning point.

If the squared bracket equals a negative number, there is no real solution. (x-4)^2=-11 cannot happen for real x. That is the same information as a negative discriminant, written in a form that also shows the minimum. Say "no real solutions" and point at the negative number. Do not write a square root of a negative.

Method: 1. Complete the square, or start from the completed form. 2. Isolate the squared bracket. 3. Square-root both sides, keeping ±. 4. Solve the two linear equations.

Read the form, then use it

Given (x-4)^2-11, you can state the turning point (4, -11) with no extra algebra. Given x^2-8x+5, you complete the square first. A question that says "by completing the square, find the minimum" wants the rewritten form on the page, not only the coordinates. The form is the method. The coordinates are the answer.

Link this to graph transformations. (x-4)^2-11 is y=x^2 translated by the vector that moves the origin to (4, -11). The inside sign and the outside constant are the same rules as f(x-4)-11. If you can complete the square, you can name the translation. If you can name the translation, you can read a completed square. They are one idea.

Method: 1. The turning point of a(x+p)^2+q is (-p, q). 2. If a>0 the turning value q is a minimum. 3. If a<0 it is a maximum. 4. Read p and q from the finished form. Do not halve again.

Turning point, minimum value and the y-intercept

Many exam questions give you a curve and say by completing the square, work out the coordinates of the turning point. For y=(x+p)^2+q the turning point is (-p, q), the line of symmetry is x=-p, and the minimum value of y is q. The y-intercept of the curve is found separately by putting x=0 into the original equation.

A minimum value question wants just the number q, while a turning point question wants both coordinates. Read which one is asked. If the coefficient of x^2 is negative, the turning point is a maximum instead, and the curve is an upside down U. Check your form by expanding it back, because a slip in the sign of p or the size of q will move the turning point.

Method: 1. Write the quadratic in the form (x+p)^2+q. 2. Read the turning point as (-p, q), changing the sign of p. 3. For a minimum value, give q only. For a line of symmetry, give x=-p. 4. Expand your answer to check it matches the original quadratic.

Tip: The sign inside the bracket flips for the coordinate and the sign outside stays.

Finding unknown constants when the form is given

Some questions give the completed-square form with unknown letters, such as x^2-6x+1=(x-a)^2-b for all values of x. Expand the bracket on the right and compare it with the left. The x term gives a, and the constant term gives b. That is faster than completing the square from scratch when you only need two numbers.

Another style gives the turning point and asks for the equation. If the turning point of y=x^2+px+q is (-3, -5), then y=(x+3)^2-5. Expand to get y=x^2+6x+4, so p=6 and q=4. Always show the expansion line. It is where the method mark is awarded, and it lets you check the constants. The method works backwards from a picture of the curve as well, so practise it both ways.

Method: 1. Expand the given form fully. 2. Match the coefficient of x on both sides to find the first letter. 3. Match the constant terms to find the second letter. 4. Substitute back and check by expanding once more.

Odd coefficients and a negative x squared

When the coefficient of x is odd, half of it is a fraction. For x^2+3x+1, half of 3 is tfrac32, so the form is (x+tfrac32)^2-tfrac94+1, which simplifies to (x+tfrac32)^2-tfrac54. Keep the fractions exact rather than rounding to a decimal. Fractions look awkward but they follow the same steps, and a calculator is not needed or wanted for this kind of question.

A negative x^2 coefficient needs one more step. Take out the factor first, as in -x^2+6x+1=-(x^2-6x)+1. Then complete the square inside the bracket and watch the minus sign when you remove the correction term. The answer is -(x-3)^2+10, and because of the minus sign the turning point (3, 10) is a maximum.

Method: 1. Take out any factor of x^2 from the x^2 and x terms only. 2. Halve the coefficient of x, keeping any fraction exact. 3. Square it and subtract that amount inside the bracket, remembering to multiply by the factor outside. 4. Add the original constant and simplify.

Using the form to solve exactly and to prove

Completing the square solves equations that do not factorise. Rearrange x^2+8x-3=0 to (x+4)^2-19=0, so (x+4)^2=19 and x=-4±√(19). The plus or minus sign is essential. If the question says give your answers in surd form, stop there and do not use a calculator. Showing every line matters, because the marks are for the completed square, for the square root step and for the final pair of answers.

The form also proves statements about a quadratic. Since a square is never negative, (x-4)^2+4 is at least 4 for every x, so x^2-8x+20 is always positive. In the same way, if the constant after the square is positive, the quadratic has no real roots. State clearly that the squared term is greater than or equal to zero.

Method: 1. Complete the square to write the quadratic as (x+p)^2+q. 2. To solve, move q across, take the square root of both sides, and include ±. 3. To prove, say that a squared term is at least 0, so the whole expression is at least q. 4. Give the final answer in the form requested, such as a surd.

Remember

Halve the linear term, square it, and balance.
When a is not 1, factor a from the x terms first.
The turning point is (-p, q) in a(x+p)^2+q.

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