Chapter 4 of 8

All 8 chapters
  1. 1Expanding two brackets, readunderstood
  2. 2Factorising quadratics, readunderstood
  3. 3Using factorising, readunderstood
  4. 4Completing the square, readunderstood
  5. 5The quadratic formula, readunderstood
  6. 6Graphs of quadratics, readunderstood
  7. 7Quadratic sequences, readunderstood
  8. 8A line and a curve, readunderstood

M4.9.4 Chapter 4

Completing the square

I know (x + 3)² = x² + 6x + 9. How can I use a square to write x² + 6x + 2?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Write x² + 6x + 2 as (x + 3)² − 7
  • Solve a quadratic by completing the square

The expression x² + 6x is nearly a perfect square. Picture a square of side x and two strips, each 3 wide and x long. The big square is not finished: its corner is missing.

Draw it first

Complete the square

  1. Step 1 shows a square marked x² with a strip marked 3x on its right and a strip marked 3x below it. The sides are marked x and 3. A dashed gap is left in the corner. Step 2 fills the gap with a hatched square marked 9, and names the whole square (x + 3)².x²3x3xx3x3

    1 A square x by x, and two strips 3 by x, make x² + 3x + 3x = x² + 6x. A corner is missing.

  2. Step 1 shows a square marked x² with a strip marked 3x on its right and a strip marked 3x below it. The sides are marked x and 3. A dashed gap is left in the corner. Step 2 fills the gap with a hatched square marked 9, and names the whole square (x + 3)².x²3x3x9x3x3(x + 3)²

    2 The missing corner is 3 by 3, with area 9. Add it, and the whole square has side x + 3.

So x² + 6x + 9 = (x + 3)². The number to add is 9. It is the square of half the x number: half of 6 is 3, and 3 × 3 = 9. A square like this is a perfect square.

To complete the square on x² + 6x + 2, make the square and then take away the extra. We added 9 to get a square, so we must take 9 away again. The value stays the same.

Worked example

Write x² + 6x + 2 in the form (x + a)² + b.

  1. The x number is 6. Half of 6 is 3.
  2. So start with the square (x + 3)². Expanding it gives x² + 6x + 9, and 3 × 3 = 9.
  3. Take away 9 and put 2 back: x² + 6x + 2 = (x + 3)² − 9 + 2.
  4. Work out −9 + 2 = −7.
  5. So x² + 6x + 2 = (x + 3)² − 7.

Answer: (x + 3)² − 7

This form shows something useful. A square is never below 0. So (x + 3)² − 7 is never below −7, and it equals −7 when x = −3. You can see it on the graph.

A graph with the x axis x and the y axis y. A curve named y = x² + 6x + 2 has its lowest point T at (−3, −7), marked with dashed guides to both axes.Ty = x² + 6x + 2−7−6−5−4−3−2−11−8−7−6−5−42340xy
The lowest point T is at x = −3, and its height is −7. The form (x + 3)² − 7 gives both numbers.

Worked example

Write x² − 8x + 3 in the form (x + a)² + b.

  1. The x number is −8. Half of −8 is −4.
  2. So start with (x − 4)². Expanding it gives x² − 8x + 16, and (−4) × (−4) = 16.
  3. Take away 16 and put 3 back: x² − 8x + 3 = (x − 4)² − 16 + 3.
  4. Work out −16 + 3 = −13.
  5. So x² − 8x + 3 = (x − 4)² − 13.

Answer: (x − 4)² − 13

Now use it to solve an equation. Once the square is complete, x appears only once. So you can undo the square with a square root. But a positive number has two square roots. The sign √ means the positive one, so the two square roots of 7 are √7 and −√7.

A table with the heads x and x × x. Rows: x is 4 and x × x is 16; x is −4 and x × x is 16; x is 3 and x × x is 9; x is −3 and x × x is 9.xx × x416−41639−39
Both 4 and −4 have the square 16. So 16 has two square roots: √16 = 4 and −√16 = −4.

Worked example

Solve x² + 6x − 7 = 0 by completing the square.

  1. Write x² + 6x − 7 as (x + 3)² − 9 − 7, which is (x + 3)² − 16.
  2. The equation is (x + 3)² − 16 = 0. Add 16 to both sides: (x + 3)² = 16.
  3. Take the square root of both sides. There are two square roots, 4 and −4: x + 3 = 4 or x + 3 = −4.
  4. Take 3 from each side: x = 1 or x = −7.
  5. Check x = −7: (−7)² + 6 × (−7) − 7 = 49 − 42 − 7 = 0.

Answer: x = 1 or x = −7

This equation also factorises: (x + 7)(x − 1) = 0 gives the same answers. Completing the square matters most when the quadratic does not factorise.

Worked example

Solve x² + 6x + 2 = 0. Give the answers to 2 decimal places.

  1. Write x² + 6x + 2 as (x + 3)² − 7. The equation is (x + 3)² − 7 = 0.
  2. Add 7 to both sides: (x + 3)² = 7.
  3. Take the square root of both sides: x + 3 = √7 or x + 3 = −√7.
  4. Use √7 ≈ 2.6458. Then x ≈ 2.6458 − 3 = −0.35, or x ≈ −2.6458 − 3 = −5.65.
  5. Check x = −0.35 in the equation. It gives 0.0225, which is close to 0, because −0.35 is rounded.

Answer: x ≈ −0.35 or x ≈ −5.65

See it three ways

With things

Cut a square of side x and two strips 5 wide. A 5 by 5 square is missing from the corner.

As a picture

A square marked x² with a strip marked 5x on its right and a strip marked 5x below it. The sides are marked x and 5. A hatched corner square is marked 25, and the whole square is named (x + 5)².x²5x5x25x5x5(x + 5)²

In words

Half of the x number 10 is 5. The missing corner has area 5 × 5 = 25. So the square is (x + 5)², and 25 must be taken away again.

In symbols

x² + 10x = (x + 5)² − 25.

Try it

Test the new form with numbers

You need: paper and a pencil.

  1. Put x = 0 into x² + 6x + 2 and into (x + 3)² − 7.
  2. Do the same for x = 1 and x = 2.
  3. Write the results in a table.

What you will see: Both forms give the same number for every x. A completed square is the same expression written another way.

If you cannot try it, this is what happens.
A table with the heads x, before and after. Before is x² + 6x + 2 and after is (x + 3)² − 7. Rows: x is 0 with 2 and 2; x is 1 with 9 and 9; x is 2 with 18 and 18.xbeforeafter02219921818

A wrong idea many people have

(x + 3)² = 16 gives x + 3 = 4, so x = 1.

Why it feels right: The square root of 16 is 4, so it seems that there is only one answer.

The test that breaks it: Put x = −7. Then (−7 + 3)² = (−4)² = 16. So x = −7 works too.

What is true: 16 has two square roots, 4 and −4. Write x + 3 = 4 or x + 3 = −4, so x = 1 or x = −7.

A wrong idea many people have

x² + 6x = (x + 3)².

Why it feels right: The 3 is half of 6, and the square looks like the start of the expression.

The test that breaks it: Put x = 1. Then x² + 6x = 7, but (1 + 3)² = 16. They are not equal.

What is true: (x + 3)² has an extra 9 that x² + 6x does not have. So x² + 6x = (x + 3)² − 9.

Key words

perfect square
an expression that is something squared, such as x² + 6x + 9, which is (x + 3)².
completing the square
writing x² + bx + c as a square plus a number.

Back to your guess

The answer is C: (x + 3)² − 7

Half of the x number 6 is 3, so the square is (x + 3)². It expands to x² + 6x + 9, which has 9 where we want 2. Take 9 away and put 2 back: 2 − 9 = −7. So x² + 6x + 2 = (x + 3)² − 7.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The picture shows x² + 12x with a gap in the corner. What number must you add to make a perfect square?

    A square marked x² with a strip marked 6x on its right and a strip marked 6x below it. The sides are marked x and 6. The corner is a dashed gap.x²6x6xx6x6
    Tap your answer.

    1. Not this one. That is the x number. Half of 12 is 6, and the number to add is 6 × 6 = 36. Go to M4.9.4. Try another.

    2. Not this one. Half of the x number is 6, but the missing corner is a square, 6 × 6 = 36. Go to M4.9.4. Try another.

    3. Yes, that is right. Half of 12 is 6, and 6 × 6 = 36. Then x² + 12x + 36 = (x + 6)².

    4. Not this one. That is 12 × 12. The number to add is the square of half of 12, which is 36. Go to M4.9.4. Try another.

    See the answer

    36

    Half of 12 is 6, and 6 × 6 = 36. Then x² + 12x + 36 = (x + 6)².

    • If you chose A: That is the x number. Half of 12 is 6, and the number to add is 6 × 6 = 36. Go to M4.9.4.
    • If you chose B: Half of the x number is 6, but the missing corner is a square, 6 × 6 = 36. Go to M4.9.4.
    • If you chose D: That is 12 × 12. The number to add is the square of half of 12, which is 36. Go to M4.9.4.

  2. Write x² + 8x + 1 in the form (x + a)² + b.

    Tap your answer.

    1. Not this one. Expanding (x + 4)² gives an extra 16, so you must take 16 away: 1 − 16 = −15. Go to M4.9.4. Try another.

    2. Not this one. Expand (x + 8)²: it gives 16x, not 8x. The number in the bracket is half of 8, which is 4. Go to M4.9.4. Try another.

    3. Not this one. You worked out −16 − 1. The +1 keeps its sign: −16 + 1 = −15. Go to M4.9.4. Try another.

    4. Yes, that is right. Half of 8 is 4, and 4 × 4 = 16. So x² + 8x + 1 = (x + 4)² − 16 + 1 = (x + 4)² − 15.

    See the answer

    (x + 4)² − 15

    Half of 8 is 4, and 4 × 4 = 16. So x² + 8x + 1 = (x + 4)² − 16 + 1 = (x + 4)² − 15.

    • If you chose A: Expanding (x + 4)² gives an extra 16, so you must take 16 away: 1 − 16 = −15. Go to M4.9.4.
    • If you chose B: Expand (x + 8)²: it gives 16x, not 8x. The number in the bracket is half of 8, which is 4. Go to M4.9.4.
    • If you chose C: You worked out −16 − 1. The +1 keeps its sign: −16 + 1 = −15. Go to M4.9.4.

  3. Write x² − 6x + 4 in the form (x + a)² + b.

    Tap your answer.

    1. Yes, that is right. Half of −6 is −3, and (−3) × (−3) = 9. So x² − 6x + 4 = (x − 3)² − 9 + 4 = (x − 3)² − 5.

    2. Not this one. Half of −6 is −3, so the bracket is (x − 3). Check the sign of the x term. Go to M4.9.4. Try another.

    3. Not this one. Expanding (x − 3)² gives an extra 9. Take it away: 4 − 9 = −5, not +5. Go to M4.9.4. Try another.

    4. Not this one. Half of −6 is −3, not −6. Expand (x − 6)²: it gives −12x, not −6x. Go to M4.9.4. Try another.

    See the answer

    (x − 3)² − 5

    Half of −6 is −3, and (−3) × (−3) = 9. So x² − 6x + 4 = (x − 3)² − 9 + 4 = (x − 3)² − 5.

    • If you chose B: Half of −6 is −3, so the bracket is (x − 3). Check the sign of the x term. Go to M4.9.4.
    • If you chose C: Expanding (x − 3)² gives an extra 9. Take it away: 4 − 9 = −5, not +5. Go to M4.9.4.
    • If you chose D: Half of −6 is −3, not −6. Expand (x − 6)²: it gives −12x, not −6x. Go to M4.9.4.

  4. Solve (x − 2)² = 9.

    Tap your answer.

    1. Not this one. 9 has two square roots, 3 and −3. Then x − 2 = −3 gives x = −1 as well. Go to M4.9.4. Try another.

    2. Yes, that is right. Take the square root of both sides: x − 2 = 3 or x − 2 = −3. Add 2: x = 5 or x = −1.

    3. Not this one. You took x − 2 = 9, which skips the square root. First take the square root of both sides: x − 2 = 3 or x − 2 = −3. Go to M4.9.4. Try another.

    4. Not this one. Add 2 to both sides: x − 2 = 3 gives x = 3 + 2 = 5. The 2 is added, not taken away. Go to M4.9.4. Try another.

    See the answer

    x = 5 or x = −1

    Take the square root of both sides: x − 2 = 3 or x − 2 = −3. Add 2: x = 5 or x = −1.

    • If you chose A: 9 has two square roots, 3 and −3. Then x − 2 = −3 gives x = −1 as well. Go to M4.9.4.
    • If you chose C: You took x − 2 = 9, which skips the square root. First take the square root of both sides: x − 2 = 3 or x − 2 = −3. Go to M4.9.4.
    • If you chose D: Add 2 to both sides: x − 2 = 3 gives x = 3 + 2 = 5. The 2 is added, not taken away. Go to M4.9.4.

  5. Solve x² + 4x − 5 = 0 by completing the square.

    Tap your answer.

    1. Not this one. x + 2 = 3 gives x = 1, since you take 2 from both sides. And x + 2 = −3 gives x = −5. Go to M4.9.4. Try another.

    2. Not this one. You found one root. Do not forget the negative square root: x + 2 = −3 gives x = −5 as well. Go to M4.9.4. Try another.

    3. Yes, that is right. x² + 4x − 5 = (x + 2)² − 4 − 5 = (x + 2)² − 9. So (x + 2)² = 9, and x + 2 = 3 or x + 2 = −3. Then x = 1 or x = −5.

    See the answer

    x = 1 or x = −5

    x² + 4x − 5 = (x + 2)² − 4 − 5 = (x + 2)² − 9. So (x + 2)² = 9, and x + 2 = 3 or x + 2 = −3. Then x = 1 or x = −5.

    • If you chose A: x + 2 = 3 gives x = 1, since you take 2 from both sides. And x + 2 = −3 gives x = −5. Go to M4.9.4.
    • If you chose B: You found one root. Do not forget the negative square root: x + 2 = −3 gives x = −5 as well. Go to M4.9.4.

  6. What is the smallest value of (x − 4)² + 3?

    Tap your answer.

    1. Not this one. The value x = 4 is where the smallest value happens. The value there is 0 + 3, which is different. Go to M4.9.4. Try another.

    2. Yes, that is right. (x − 4)² is never below 0 and is 0 at x = 4, so the smallest value is 3.

    3. Not this one. x = 0 gives 16 + 3 = 19, but that is not the smallest. A square is never below 0. Go to M4.9.4. Try another.

    See the answer

    3

    (x − 4)² is never below 0 and is 0 at x = 4, so the smallest value is 3.

    • If you chose A: The value x = 4 is where the smallest value happens. The value there is 0 + 3, which is different. Go to M4.9.4.
    • If you chose C: x = 0 gives 16 + 3 = 19, but that is not the smallest. A square is never below 0. Go to M4.9.4.

  7. Write x² + 5x + 1 in the form (x + a)² + b. What are a and b?

    Tap your answer.

    1. Yes, that is right. Half of 5 is 2.5, and 2.5 × 2.5 = 6.25. So x² + 5x + 1 = (x + 2.5)² − 6.25 + 1, and b = −5.25.

    2. Not this one. The number a is half of the x number, so a = 2.5. Expanding (x + 5)² gives 10x, not 5x. Go to M4.9.4. Try another.

    3. Not this one. Expanding (x + 2.5)² gives an extra 6.25. Take it away: 1 − 6.25 = −5.25. Go to M4.9.4. Try another.

    See the answer

    a = 2.5, b = −5.25

    Half of 5 is 2.5, and 2.5 × 2.5 = 6.25. So x² + 5x + 1 = (x + 2.5)² − 6.25 + 1, and b = −5.25.

    • If you chose B: The number a is half of the x number, so a = 2.5. Expanding (x + 5)² gives 10x, not 5x. Go to M4.9.4.
    • If you chose C: Expanding (x + 2.5)² gives an extra 6.25. Take it away: 1 − 6.25 = −5.25. Go to M4.9.4.

  8. From an earlier chapter

    Solve x² − 10x + 21 = 0 by factorising.

    Tap your answer.

    1. Not this one. The brackets are (x − 3)(x − 7). Putting x − 3 = 0 gives x = 3, not −3. Go to M4.9.3. Try another.

    2. Not this one. These are the numbers in the equation. The solutions come from the brackets, and they are 3 and 7. Go to M4.9.2. Try another.

    3. Not this one. Both numbers in the brackets are negative, (x − 3)(x − 7), so both solutions are positive. Go to M4.9.3. Try another.

    4. Yes, that is right. Two numbers with product 21 and sum −10 are −3 and −7. So (x − 3)(x − 7) = 0, and x = 3 or x = 7.

    See the answer

    x = 3 or x = 7

    Two numbers with product 21 and sum −10 are −3 and −7. So (x − 3)(x − 7) = 0, and x = 3 or x = 7.

    • If you chose A: The brackets are (x − 3)(x − 7). Putting x − 3 = 0 gives x = 3, not −3. Go to M4.9.3.
    • If you chose B: These are the numbers in the equation. The solutions come from the brackets, and they are 3 and 7. Go to M4.9.2.
    • If you chose C: Both numbers in the brackets are negative, (x − 3)(x − 7), so both solutions are positive. Go to M4.9.3.

  9. x² + 10x + 31 = (x + a)² + b. Give a, then b.

    See the answer

    a = 5, b = 6

    Half of 10 is 5, and 5 × 5 = 25. So x² + 10x + 31 = (x + 5)² − 25 + 31 = (x + 5)² + 6.

  10. The equation (x − 3)² = 25 has two solutions. What is the larger one?

    See the answer

    x = 8

    Take the square root of both sides: x − 3 = 5 or x − 3 = −5. So x = 8 or x = −2. The larger is 8.

Keep it

Copy this into your exercise book.

Half the x number goes in the bracket. Take away its square. Then add, take the square root, and use plus and minus.

Draw: A square with two strips and a hatched corner, with x² + 6x + 9 = (x + 3)² written under it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (898 KB). It opens on any phone.