GCSE Maths: Completing the square (Higher)
Part of the GCSE Maths quadratic and simultaneous equations chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Completing the square
Look at what happens when you expand a bracket that is squared: (x + 3)2 = x2 + 6x + 9. The expression x2 + 6x + 9 is a perfect square. Completing the square means rewriting any quadratic x2 + bx + c in the form (x + p)2 + q, where the letter x appears only once. It is useful for three things: solving equations, finding the turning point of a graph, and proving things about quadratics.
Why it works. In the picture, the area x2 + 6x is an x by x square plus two strips of 3 by x. To fill the corner you need 9 more. So x2 + 6x = (x + 3)2 − 9. The rule is: halve the number in front of x, put it in the bracket, and subtract its square. So x2 + bx + c = (x + b/2)2 − (b/2)2 + c. For x2 + 6x + 2 this gives (x + 3)2 − 9 + 2 = (x + 3)2 − 7.
When the number in front of x2 is not 1, take that number out of the first two terms first: 2x2 + 8x + 5 = 2(x2 + 4x) + 5 = 2[(x + 2)2 − 4] + 5 = 2(x + 2)2 − 8 + 5 = 2(x + 2)2 − 3.
Solving. To solve x2 + 6x + 2 = 0, write (x + 3)2 − 7 = 0, so (x + 3)2 = 7, so x + 3 = ±√7, and x = −3 ± √7. Always include both the + and the − square roots.
Turning points. The curve y = (x + p)2 + q has a squared bracket, which is never negative and is 0 when x = −p. So the curve’s lowest point is (−p, q). The x-coordinate has the opposite sign to the number in the bracket. For example y = x2 − 6x + 5 = (x − 3)2 − 4 has its minimum at (3, −4).
Key ideas
- x2 + bx + c = (x + b/2)2 − (b/2)2 + c: halve b, then subtract its square.
- If the number in front of x2 is a (not 1), factor a out of the x2 and x terms first.
- To solve: rearrange to (x + p)2 = number, square-root both sides (±), then subtract p.
- y = a(x + p)2 + q has its turning point at (−p, q): a minimum if a > 0 and a maximum if a < 0.
- Check by expanding the answer.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 + 6x + 2
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.6 ÷ 2 = 3
The bracket is (x + 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)2 = x2 + 6x + 9 - Adjust the number on the end. Our expression has x2 + 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.x2 + 6x + 2 = (x + 3)2 − 9 + 2
= (x + 3)2 − 7
Check. Expand the answer: (x + 3)2 − 7 gives x2 + 6x + 2, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 − 8x + 5
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−8 ÷ 2 = −4
The bracket is (x − 4)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−4) = −4x
Inner: (−4) × x = −4x
Last: (−4) × (−4) = 16
(x − 4)2 = x2 − 8x + 16 - Adjust the number on the end. Our expression has x2 − 8x + 5, but the bracket gave an extra +16 at the end. So we must take that off again and then put on our own 5.x2 − 8x + 5 = (x − 4)2 − 16 + 5
= (x − 4)2 − 11
Check. Expand the answer: (x − 4)2 − 11 gives x2 − 8x + 5, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.2x2 + 8x + 5
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 + 8x + 5 = 2(x2 + 4x) + 5
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.4 ÷ 2 = 2
x2 + 4x = (x + 2)2 − 4 - Multiply the extra number by 2 and tidy up. The −4 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x + 2)2 − 4] + 5
= 2(x + 2)2 − 8 + 5
= 2(x + 2)2 − 3
Check. Expand the answer: 2(x + 2)2 − 3 gives 2x2 + 8x + 5, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 + 6x + 2 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.6 ÷ 2 = 3
The bracket is (x + 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)2 = x2 + 6x + 9 - Adjust the number on the end. Our expression has x2 + 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.x2 + 6x + 2 = (x + 3)2 − 9 + 2
= (x + 3)2 − 7 - Write the equation using the new form. The expression is now equal to 0.(x + 3)2 − 7 = 0
- Move the number to the right-hand side. Add 7 to both sides.(x + 3)2 = 7
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 3 = ±√7
- Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −3 ± √7
Check. Substitute x = −5.65: 1 × (−5.65)2 + 6 × (−5.65) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −5.65 is rounded). ✓
Substitute x = −0.35: 1 × (−0.35)2 + 6 × (−0.35) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −0.35 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.2x2 + 12x + 6 = 0
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 + 12x + 6 = 2(x2 + 6x) + 6
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.6 ÷ 2 = 3
x2 + 6x = (x + 3)2 − 9 - Multiply the extra number by 2 and tidy up. The −9 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x + 3)2 − 9] + 6
= 2(x + 3)2 − 18 + 6
= 2(x + 3)2 − 12 - Write the equation using the new form. The expression is now equal to 0.2(x + 3)2 − 12 = 0
- Move the number to the right-hand side. Add 12 to both sides.2(x + 3)2 = 12
- Divide both sides by 2. This leaves the bracket squared on its own.(x + 3)2 = 12 ÷ 2 = 6
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 3 = ±√6
- Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −3 ± √6
- Use a calculator to round to 2 decimal places. Work out the + and the − separately.x = −3 + √6 = −0.5505… ≈ −0.55
x = −3 − √6 = −5.4495… ≈ −5.45
Check. Substitute x = −5.45: 2 × (−5.45)2 + 12 × (−5.45) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −5.45 is rounded). ✓
Substitute x = −0.55: 2 × (−0.55)2 + 12 × (−0.55) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −0.55 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 − 4x + 7 can be read straight from the completed-square form.x2 − 4x + 7
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−4 ÷ 2 = −2
The bracket is (x − 2)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−2) = −2x
Inner: (−2) × x = −2x
Last: (−2) × (−2) = 4
(x − 2)2 = x2 − 4x + 4 - Adjust the number on the end. Our expression has x2 − 4x + 7, but the bracket gave an extra +4 at the end. So we must take that off again and then put on our own 7.x2 − 4x + 7 = (x − 2)2 − 4 + 7
= (x − 2)2 + 3 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 2 = 0, i.e. x = 2. At that point the whole expression equals the number on the end, 3. Every other value of the bracket makes the expression bigger, so this is the minimum.(x − 2)2 + 3
Turning point = (2, 3), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = 2. At x = 2: y = 22 − 4 × 2 + 7 = 4 − 8 + 7 = 3. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Watch out
When you complete the square you subtract the square of half the number (not add it), and when you read off the turning point from (x + 3)2 − 7 the x-coordinate is −3, not 3. The number outside the bracket is q, with its own sign: (x + 3)2 − 7 has y-coordinate −7.Exercise 14J Completing the square
Non-calculatorKey words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 + 4x + 1
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.4 ÷ 2 = 2
The bracket is (x + 2)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 2 = 2x
Inner: 2 × x = 2x
Last: 2 × 2 = 4
(x + 2)2 = x2 + 4x + 4 - Adjust the number on the end. Our expression has x2 + 4x + 1, but the bracket gave an extra +4 at the end. So we must take that off again and then put on our own 1.x2 + 4x + 1 = (x + 2)2 − 4 + 1
= (x + 2)2 − 3
Check. Expand the answer: (x + 2)2 − 3 gives x2 + 4x + 1, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 + 10x + 3
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.10 ÷ 2 = 5
The bracket is (x + 5)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 5 = 5x
Inner: 5 × x = 5x
Last: 5 × 5 = 25
(x + 5)2 = x2 + 10x + 25 - Adjust the number on the end. Our expression has x2 + 10x + 3, but the bracket gave an extra +25 at the end. So we must take that off again and then put on our own 3.x2 + 10x + 3 = (x + 5)2 − 25 + 3
= (x + 5)2 − 22
Check. Expand the answer: (x + 5)2 − 22 gives x2 + 10x + 3, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 − 6x + 2
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−6 ÷ 2 = −3
The bracket is (x − 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−3) = −3x
Inner: (−3) × x = −3x
Last: (−3) × (−3) = 9
(x − 3)2 = x2 − 6x + 9 - Adjust the number on the end. Our expression has x2 − 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.x2 − 6x + 2 = (x − 3)2 − 9 + 2
= (x − 3)2 − 7
Check. Expand the answer: (x − 3)2 − 7 gives x2 − 6x + 2, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 − 2x − 5
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−2 ÷ 2 = −1
The bracket is (x − 1)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−1) = −x
Inner: (−1) × x = −x
Last: (−1) × (−1) = 1
(x − 1)2 = x2 − 2x + 1 - Adjust the number on the end. Our expression has x2 − 2x − 5, but the bracket gave an extra +1 at the end. So we must take that off again and then put on our own −5.x2 − 2x − 5 = (x − 1)2 − 1 − 5
= (x − 1)2 − 6
Check. Expand the answer: (x − 1)2 − 6 gives x2 − 2x − 5, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 + 3x + 1
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.3 ÷ 2 = 32
The bracket is (x + 32)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 3/2 = 3/2x
Inner: 3/2 × x = 3/2x
Last: 3/2 × 3/2 = 9/4
(x + 32)2 = x2 + 3x + 9/4 - Adjust the number on the end. Our expression has x2 + 3x + 1, but the bracket gave an extra +94 at the end. So we must take that off again and then put on our own 1.x2 + 3x + 1 = (x + 32)2 − 94 + 1
= (x + 32)2 − 54
Check. Expand the answer: (x + 32)2 − 54 gives x2 + 3x + 1, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.x2 − 12x + 40
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−12 ÷ 2 = −6
The bracket is (x − 6)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−6) = −6x
Inner: (−6) × x = −6x
Last: (−6) × (−6) = 36
(x − 6)2 = x2 − 12x + 36 - Adjust the number on the end. Our expression has x2 − 12x + 40, but the bracket gave an extra +36 at the end. So we must take that off again and then put on our own 40.x2 − 12x + 40 = (x − 6)2 − 36 + 40
= (x − 6)2 + 4
Check. Expand the answer: (x − 6)2 + 4 gives x2 − 12x + 40, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.2x2 + 8x + 3
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 + 8x + 3 = 2(x2 + 4x) + 3
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.4 ÷ 2 = 2
x2 + 4x = (x + 2)2 − 4 - Multiply the extra number by 2 and tidy up. The −4 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x + 2)2 − 4] + 3
= 2(x + 2)2 − 8 + 3
= 2(x + 2)2 − 5
Check. Expand the answer: 2(x + 2)2 − 5 gives 2x2 + 8x + 3, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.3x2 − 12x + 5
- Take the number 3 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.3x2 − 12x + 5 = 3(x2 − 4x) + 5
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−4 ÷ 2 = −2
x2 − 4x = (x − 2)2 − 4 - Multiply the extra number by 3 and tidy up. The −4 is still inside the bracket that is multiplied by 3, so it must be multiplied by 3 to come outside. Then add the number on the end.3[(x − 2)2 − 4] + 5
= 3(x − 2)2 − 12 + 5
= 3(x − 2)2 − 7
Check. Expand the answer: 3(x − 2)2 − 7 gives 3x2 − 12x + 5, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.2x2 − 4x − 1
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 − 4x − 1 = 2(x2 − 2x) − 1
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−2 ÷ 2 = −1
x2 − 2x = (x − 1)2 − 1 - Multiply the extra number by 2 and tidy up. The −1 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x − 1)2 − 1] − 1
= 2(x − 1)2 − 2 − 1
= 2(x − 1)2 − 3
Check. Expand the answer: 2(x − 1)2 − 3 gives 2x2 − 4x − 1, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.5x2 + 10x + 2
- Take the number 5 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.5x2 + 10x + 2 = 5(x2 + 2x) + 2
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.2 ÷ 2 = 1
x2 + 2x = (x + 1)2 − 1 - Multiply the extra number by 5 and tidy up. The −1 is still inside the bracket that is multiplied by 5, so it must be multiplied by 5 to come outside. Then add the number on the end.5[(x + 1)2 − 1] + 2
= 5(x + 1)2 − 5 + 2
= 5(x + 1)2 − 3
Check. Expand the answer: 5(x + 1)2 − 3 gives 5x2 + 10x + 2, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.4x2 + 8x + 9
- Take the number 4 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.4x2 + 8x + 9 = 4(x2 + 2x) + 9
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.2 ÷ 2 = 1
x2 + 2x = (x + 1)2 − 1 - Multiply the extra number by 4 and tidy up. The −1 is still inside the bracket that is multiplied by 4, so it must be multiplied by 4 to come outside. Then add the number on the end.4[(x + 1)2 − 1] + 9
= 4(x + 1)2 − 4 + 9
= 4(x + 1)2 + 5
Check. Expand the answer: 4(x + 1)2 + 5 gives 4x2 + 8x + 9, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q.
Method. Completing the square: take out a if it is not 1, halve the number in front of x to get p, then correct the number on the end.
- Write down the expression and what we want. We want to write it in the form a(x + p)2 + q, with x appearing only once.−x2 + 6x + 1
- Take the number −1 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.−x2 + 6x + 1 = −(x2 − 6x) + 1
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−6 ÷ 2 = −3
x2 − 6x = (x − 3)2 − 9 - Multiply the extra number by −1 and tidy up. The −9 is still inside the bracket that is multiplied by −1, so it must be multiplied by −1 to come outside. Then add the number on the end.−[(x − 3)2 − 9] + 1
= −(x − 3)2 + 9 + 1
= −(x − 3)2 + 10
Check. Expand the answer: −(x − 3)2 + 10 gives −x2 + 6x + 1, the same as the original. ✓
Watch out. Halve the number in front of x, and then subtract its square. Do not add it.
x2 + 4x − 1 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 + 4x − 1 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.4 ÷ 2 = 2
The bracket is (x + 2)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 2 = 2x
Inner: 2 × x = 2x
Last: 2 × 2 = 4
(x + 2)2 = x2 + 4x + 4 - Adjust the number on the end. Our expression has x2 + 4x − 1, but the bracket gave an extra +4 at the end. So we must take that off again and then put on our own −1.x2 + 4x − 1 = (x + 2)2 − 4 − 1
= (x + 2)2 − 5 - Write the equation using the new form. The expression is now equal to 0.(x + 2)2 − 5 = 0
- Move the number to the right-hand side. Add 5 to both sides.(x + 2)2 = 5
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 2 = ±√5
- Subtract 2 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −2 ± √5
Check. Substitute x = −4.24: 1 × (−4.24)2 + 4 × (−4.24) + (−1) ≈ 0.018, which is very close to 0 (not exactly 0 only because −4.24 is rounded). ✓
Substitute x = 0.24: 1 × (0.24)2 + 4 × (0.24) + (−1) ≈ 0.018, which is very close to 0 (not exactly 0 only because 0.24 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
x2 − 6x + 1 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 − 6x + 1 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−6 ÷ 2 = −3
The bracket is (x − 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−3) = −3x
Inner: (−3) × x = −3x
Last: (−3) × (−3) = 9
(x − 3)2 = x2 − 6x + 9 - Adjust the number on the end. Our expression has x2 − 6x + 1, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 1.x2 − 6x + 1 = (x − 3)2 − 9 + 1
= (x − 3)2 − 8 - Write the equation using the new form. The expression is now equal to 0.(x − 3)2 − 8 = 0
- Move the number to the right-hand side. Add 8 to both sides.(x − 3)2 = 8
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x − 3 = ±√8
- Simplify √8. 8 = 4 × 2, and 4 is a square number, so √8 = 2√2.x − 3 = ±2√2
- Add 3 to both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = 3 ± 2√2
Check. Substitute x = 0.17: 1 × (0.17)2 + (−6) × (0.17) + 1 ≈ 0.009, which is very close to 0 (not exactly 0 only because 0.17 is rounded). ✓
Substitute x = 5.83: 1 × (5.83)2 + (−6) × (5.83) + 1 ≈ 0.009, which is very close to 0 (not exactly 0 only because 5.83 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
x2 + 8x + 3 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 + 8x + 3 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.8 ÷ 2 = 4
The bracket is (x + 4)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 4 = 4x
Inner: 4 × x = 4x
Last: 4 × 4 = 16
(x + 4)2 = x2 + 8x + 16 - Adjust the number on the end. Our expression has x2 + 8x + 3, but the bracket gave an extra +16 at the end. So we must take that off again and then put on our own 3.x2 + 8x + 3 = (x + 4)2 − 16 + 3
= (x + 4)2 − 13 - Write the equation using the new form. The expression is now equal to 0.(x + 4)2 − 13 = 0
- Move the number to the right-hand side. Add 13 to both sides.(x + 4)2 = 13
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 4 = ±√13
- Subtract 4 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −4 ± √13
Check. Substitute x = −7.61: 1 × (−7.61)2 + 8 × (−7.61) + 3 ≈ 0.032, which is very close to 0 (not exactly 0 only because −7.61 is rounded). ✓
Substitute x = −0.39: 1 × (−0.39)2 + 8 × (−0.39) + 3 ≈ 0.032, which is very close to 0 (not exactly 0 only because −0.39 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
x2 − 2x − 5 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 − 2x − 5 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−2 ÷ 2 = −1
The bracket is (x − 1)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−1) = −x
Inner: (−1) × x = −x
Last: (−1) × (−1) = 1
(x − 1)2 = x2 − 2x + 1 - Adjust the number on the end. Our expression has x2 − 2x − 5, but the bracket gave an extra +1 at the end. So we must take that off again and then put on our own −5.x2 − 2x − 5 = (x − 1)2 − 1 − 5
= (x − 1)2 − 6 - Write the equation using the new form. The expression is now equal to 0.(x − 1)2 − 6 = 0
- Move the number to the right-hand side. Add 6 to both sides.(x − 1)2 = 6
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x − 1 = ±√6
- Add 1 to both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = 1 ± √6
Check. Substitute x = −1.45: 1 × (−1.45)2 + (−2) × (−1.45) + (−5) ≈ 0.003, which is very close to 0 (not exactly 0 only because −1.45 is rounded). ✓
Substitute x = 3.45: 1 × (3.45)2 + (−2) × (3.45) + (−5) ≈ 0.003, which is very close to 0 (not exactly 0 only because 3.45 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
x2 + 6x + 2 = 0. Give your answers correct to 2 decimal places.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.x2 + 6x + 2 = 0
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.6 ÷ 2 = 3
The bracket is (x + 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)2 = x2 + 6x + 9 - Adjust the number on the end. Our expression has x2 + 6x + 2, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 2.x2 + 6x + 2 = (x + 3)2 − 9 + 2
= (x + 3)2 − 7 - Write the equation using the new form. The expression is now equal to 0.(x + 3)2 − 7 = 0
- Move the number to the right-hand side. Add 7 to both sides.(x + 3)2 = 7
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 3 = ±√7
- Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −3 ± √7
- Use a calculator to round to 2 decimal places. Work out the + and the − separately.x = −3 + √7 = −0.3542… ≈ −0.35
x = −3 − √7 = −5.6458… ≈ −5.65
Check. Substitute x = −5.65: 1 × (−5.65)2 + 6 × (−5.65) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −5.65 is rounded). ✓
Substitute x = −0.35: 1 × (−0.35)2 + 6 × (−0.35) + 2 ≈ 0.023, which is very close to 0 (not exactly 0 only because −0.35 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
2x2 + 12x + 6 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.2x2 + 12x + 6 = 0
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 + 12x + 6 = 2(x2 + 6x) + 6
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.6 ÷ 2 = 3
x2 + 6x = (x + 3)2 − 9 - Multiply the extra number by 2 and tidy up. The −9 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x + 3)2 − 9] + 6
= 2(x + 3)2 − 18 + 6
= 2(x + 3)2 − 12 - Write the equation using the new form. The expression is now equal to 0.2(x + 3)2 − 12 = 0
- Move the number to the right-hand side. Add 12 to both sides.2(x + 3)2 = 12
- Divide both sides by 2. This leaves the bracket squared on its own.(x + 3)2 = 12 ÷ 2 = 6
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x + 3 = ±√6
- Subtract 3 from both sides. This leaves x on its own. This is the exact answer: it keeps the surd.x = −3 ± √6
Check. Substitute x = −5.45: 2 × (−5.45)2 + 12 × (−5.45) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −5.45 is rounded). ✓
Substitute x = −0.55: 2 × (−0.55)2 + 12 × (−0.55) + 6 ≈ 0.005, which is very close to 0 (not exactly 0 only because −0.55 is rounded). ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
3x2 − 18x + 15 = 0. Give your answers in the form p ± √q.
Key words. Completing the square means rewriting x2 + bx + c as (x + p)2 + q. A surd is a square root that is not a whole number.
Method. Completing the square: write the quadratic as a(x + p)2 + q, then rearrange to find x by square-rooting.
- Write the equation. We will rewrite the left-hand side so that x appears only once, inside a bracket that is squared.3x2 − 18x + 15 = 0
- Take the number 3 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.3x2 − 18x + 15 = 3(x2 − 6x) + 15
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−6 ÷ 2 = −3
x2 − 6x = (x − 3)2 − 9 - Multiply the extra number by 3 and tidy up. The −9 is still inside the bracket that is multiplied by 3, so it must be multiplied by 3 to come outside. Then add the number on the end.3[(x − 3)2 − 9] + 15
= 3(x − 3)2 − 27 + 15
= 3(x − 3)2 − 12 - Write the equation using the new form. The expression is now equal to 0.3(x − 3)2 − 12 = 0
- Move the number to the right-hand side. Add 12 to both sides.3(x − 3)2 = 12
- Divide both sides by 3. This leaves the bracket squared on its own.(x − 3)2 = 12 ÷ 3 = 4
- Take the square root of both sides. There are two square roots of a positive number: a positive one and a negative one. For example 3 and −3 both square to make 9. So write ±.x − 3 = ±√4
- Work out the square root of 4. 2 × 2 = 4x − 3 = ±2
- Add 3 to both sides. This leaves x on its own. Work out both answers.x = 3 + 2 = 5 or x = 3 − 2 = 1
Check. When x = 1: 3 × 12 − 18 × 1 + 15 = 3 − 18 + 15 = 0. ✓
When x = 5: 3 × 52 − 18 × 5 + 15 = 75 − 90 + 15 = 0. ✓
Watch out. Square-rooting gives two answers (±). And the number inside the bracket is half of the number in front of x: remember to halve it, and to use the opposite sign when you solve.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 + 6x + 5 can be read straight from the completed-square form.x2 + 6x + 5
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.6 ÷ 2 = 3
The bracket is (x + 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 3 = 3x
Inner: 3 × x = 3x
Last: 3 × 3 = 9
(x + 3)2 = x2 + 6x + 9 - Adjust the number on the end. Our expression has x2 + 6x + 5, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 5.x2 + 6x + 5 = (x + 3)2 − 9 + 5
= (x + 3)2 − 4 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x + 3 = 0, i.e. x = −3. At that point the whole expression equals the number on the end, −4. Every other value of the bracket makes the expression bigger, so this is the minimum.(x + 3)2 − 4
Turning point = (−3, −4), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = −3. At x = −3: y = (−3)2 + 6 × (−3) + 5 = 9 − 18 + 5 = −4. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 − 4x + 7 can be read straight from the completed-square form.x2 − 4x + 7
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−4 ÷ 2 = −2
The bracket is (x − 2)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−2) = −2x
Inner: (−2) × x = −2x
Last: (−2) × (−2) = 4
(x − 2)2 = x2 − 4x + 4 - Adjust the number on the end. Our expression has x2 − 4x + 7, but the bracket gave an extra +4 at the end. So we must take that off again and then put on our own 7.x2 − 4x + 7 = (x − 2)2 − 4 + 7
= (x − 2)2 + 3 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 2 = 0, i.e. x = 2. At that point the whole expression equals the number on the end, 3. Every other value of the bracket makes the expression bigger, so this is the minimum.(x − 2)2 + 3
Turning point = (2, 3), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = 2. At x = 2: y = 22 − 4 × 2 + 7 = 4 − 8 + 7 = 3. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 + 2x − 8 can be read straight from the completed-square form.x2 + 2x − 8
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.2 ÷ 2 = 1
The bracket is (x + 1)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
(x + 1)2 = x2 + 2x + 1 - Adjust the number on the end. Our expression has x2 + 2x − 8, but the bracket gave an extra +1 at the end. So we must take that off again and then put on our own −8.x2 + 2x − 8 = (x + 1)2 − 1 − 8
= (x + 1)2 − 9 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x + 1 = 0, i.e. x = −1. At that point the whole expression equals the number on the end, −9. Every other value of the bracket makes the expression bigger, so this is the minimum.(x + 1)2 − 9
Turning point = (−1, −9), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = −1. At x = −1: y = (−1)2 + 2 × (−1) − 8 = 1 − 2 − 8 = −9. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = 2x2 − 8x + 9 can be read straight from the completed-square form.2x2 − 8x + 9
- Take the number 2 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.2x2 − 8x + 9 = 2(x2 − 4x) + 9
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−4 ÷ 2 = −2
x2 − 4x = (x − 2)2 − 4 - Multiply the extra number by 2 and tidy up. The −4 is still inside the bracket that is multiplied by 2, so it must be multiplied by 2 to come outside. Then add the number on the end.2[(x − 2)2 − 4] + 9
= 2(x − 2)2 − 8 + 9
= 2(x − 2)2 + 1 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 2 = 0, i.e. x = 2. At that point the whole expression equals the number on the end, 1. Every other value of the bracket makes the expression bigger, so this is the minimum.2(x − 2)2 + 1
Turning point = (2, 1), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = 2. At x = 2: y = 2 × 22 − 8 × 2 + 9 = 8 − 16 + 9 = 1. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 − 10x can be read straight from the completed-square form.x2 − 10x
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−10 ÷ 2 = −5
The bracket is (x − 5)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−5) = −5x
Inner: (−5) × x = −5x
Last: (−5) × (−5) = 25
(x − 5)2 = x2 − 10x + 25 - Adjust the number on the end. Our expression has x2 − 10x, but the bracket gave an extra +25 at the end. So we must take that off again and then put on our own 0.x2 − 10x = (x − 5)2 − 25 + 0
= (x − 5)2 − 25 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 5 = 0, i.e. x = 5. At that point the whole expression equals the number on the end, −25. Every other value of the bracket makes the expression bigger, so this is the minimum.(x − 5)2 − 25
Turning point = (5, −25), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = 5. At x = 5: y = 52 − 10 × 5 = 25 − 50 = −25. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = −x2 + 6x + 1 can be read straight from the completed-square form.−x2 + 6x + 1
- Take the number −1 out of the x2 and x terms only. We can only complete the square when the number in front of x2 is 1. So factorise just the first two terms, and leave the number on the end alone.−x2 + 6x + 1 = −(x2 − 6x) + 1
- Complete the square inside the bracket. Halve the number in front of x inside the bracket. Then subtract its square, because squaring the bracket adds it.−6 ÷ 2 = −3
x2 − 6x = (x − 3)2 − 9 - Multiply the extra number by −1 and tidy up. The −9 is still inside the bracket that is multiplied by −1, so it must be multiplied by −1 to come outside. Then add the number on the end.−[(x − 3)2 − 9] + 1
= −(x − 3)2 + 9 + 1
= −(x − 3)2 + 10 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 3 = 0, i.e. x = 3. At that point the whole expression equals the number on the end, 10. Every other value of the bracket makes the expression smaller (the curve is upside down because the number in front of the bracket is negative), so this is the maximum.−(x − 3)2 + 10
Turning point = (3, 10), a maximum.
Check. Check using the graph: the curve should be symmetrical about x = 3. At x = 3: y = −32 + 6 × 3 + 1 = −9 + 18 + 1 = 10. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Key words. The turning point is the lowest point (minimum) or highest point (maximum) of a quadratic curve.
Method. Complete the square: y = a(x + p)2 + q has its turning point at (−p, q).
- Write down the expression. The turning point of the graph y = x2 − 6x + 15 can be read straight from the completed-square form.x2 − 6x + 15
- Halve the number in front of x. The bracket we want is (x + something)2. When you expand (x + p)2 you get x2 + 2px + p2, so the number in front of x is twice p. So p is half of it.−6 ÷ 2 = −3
The bracket is (x − 3)2. - Expand the bracket to see what it really gives. The bracket gives the right x2 and x terms, but it also gives an extra number that is not in our expression.First: x × x = x2
Outer: x × (−3) = −3x
Inner: (−3) × x = −3x
Last: (−3) × (−3) = 9
(x − 3)2 = x2 − 6x + 9 - Adjust the number on the end. Our expression has x2 − 6x + 15, but the bracket gave an extra +9 at the end. So we must take that off again and then put on our own 15.x2 − 6x + 15 = (x − 3)2 − 9 + 15
= (x − 3)2 + 6 - Read off the turning point. A squared bracket is never negative, and it is exactly 0 when the bracket is 0. That happens when x − 3 = 0, i.e. x = 3. At that point the whole expression equals the number on the end, 6. Every other value of the bracket makes the expression bigger, so this is the minimum.(x − 3)2 + 6
Turning point = (3, 6), a minimum.
Check. Check using the graph: the curve should be symmetrical about x = 3. At x = 3: y = 32 − 6 × 3 + 15 = 9 − 18 + 15 = 6. ✓
Watch out. The x-coordinate of the turning point has the opposite sign to the number inside the bracket: (x + 3)2 gives x = −3.
Method. Complete the square, then use the fact that a squared number is never negative.
- Complete the square. Halve 6 to get 3, and subtract 32 = 9.x2 + 6x + 11 = (x + 3)2 − 9 + 11 = (x + 3)2 + 2
- Use the squared bracket. Whatever x is, (x + 3)2 is a square, so it is at least 0.(x + 3)2 ≥ 0
- Add 2. Adding 2 to a number that is at least 0 gives a number that is at least 2.(x + 3)2 + 2 ≥ 2
- Conclude. 2 is positive.So x2 + 6x + 11 is always positive.
Check. Test x = −3 (where the bracket is 0): 9 − 18 + 11 = 2, which is positive and is the smallest value. ✓
Watch out. In a proof, say why a bracket squared is never negative.
Method. Expand the completed-square form and compare it with the given expression, term by term.
- Expand (x + p)2. Use the rule (x + p)2 = x2 + 2px + p2.(x + p)2 + q = x2 + 2px + p2 + q
- Compare the x terms. The number in front of x must match: 2p = −10.p = −5
- Compare the numbers. p2 + q must equal 30.25 + q = 30, so q = 5
Check. (x − 5)2 + 5 = x2 − 10x + 25 + 5 = x2 − 10x + 30. ✓
Watch out. p is half of the number in front of x, with the same sign.
Method. Work backwards from the turning point. A curve with turning point (3, −2) and the form y = x2 + bx + c is y = (x − 3)2 − 2.
- Write the completed-square form. The turning point (−p, q) = (3, −2), so p = −3 and q = −2.y = (x − 3)2 − 2
- Expand the bracket. First: x × x = x2
Outer: x × (−3) = −3x
Inner: (−3) × x = −3x
Last: (−3) × (−3) = 9
(x − 3)2 = x2 − 6x + 9 - Collect the numbers. Add the −2.y = x2 − 6x + 9 − 2 = x2 − 6x + 7
- Read off b and c. Compare with y = x2 + bx + c.b = −6, c = 7
Check. Complete the square on x2 − 6x + 7: (x − 3)2 − 9 + 7 = (x − 3)2 − 2, with turning point (3, −2). ✓
Watch out. Expanding is the reverse of completing the square.
Method. The equation x2 + 6x + 11 = 0 has solutions only if the curve y = x2 + 6x + 11 reaches the x-axis. Use its minimum value.
- Complete the square. This gives the turning point.x2 + 6x + 11 = (x + 3)2 + 2
Turning point: (−3, 2) - Think about the graph. The curve opens upwards and its lowest point has y = 2, above the x-axis. So the curve never reaches y = 0.No point on the curve has y = 0.
- Conclude. No x makes the expression equal to 0.The equation has no real solutions.
Check. Using the discriminant: 62 − 4 × 1 × 11 = 36 − 44 = −8, which is negative. ✓
Watch out. A curve that opens upwards and has a positive minimum value never meets the x-axis.