Quadratics. Attnom Academy

M4.9.1 Chapter 1

Expanding two brackets

I know 3(x + 4) = 3x + 12. What do I do when both parts are brackets, like (x + 3)(x + 2)?

Your guess

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

Your guess

A x² + 6: multiply the first terms, and multiply the last terms.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x² + 5x + 6: multiply every term by every term, then join the like terms.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 2x + 5: add the terms inside the brackets.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Expand (x + 3)(x − 5)
Expand (2x + 1)² without losing the middle term

Amina knows that 3(x + 4) means 3 × x and 3 × 4. The 3 multiplies each part of the bracket. This is called expanding the bracket.

Now take (x + 3)(x + 2). Picture a rectangle that is x + 3 long and x + 2 wide. Its area is the two brackets multiplied together.

The whole rectangle is made of four parts. Its area is x² + 3x + 2x + 6.

The four parts are the four products. x times x is x². x times 2 is 2x. 3 times x is 3x. 3 times 2 is 6. The parts 3x and 2x are alike, so they join: 5x.

Draw it first

Multiply every pair of terms

1 Start with the two brackets, (x + 3)(x + 2).
2 Multiply every term in the first bracket by every term in the second. There are 2 × 2 = 4 products.
3 Join the like terms. 2x + 3x = 5x, so the answer is x² + 5x + 6.

A term with a minus sign needs care. A positive number times a negative number is negative. A negative times a negative is positive. Here is a pattern that shows why.

Going down the table, the second number falls by 1 each time and the answer rises by 5. So −5 × −1 is 5.

A grid keeps the four products in order. Write one bracket across the top and the other down the side. Each cell holds the product of the terms at the ends of its row and column.

The hatched cells are the like terms. Adding them gives 3x − 5x = −2x.

Worked example

Expand (x + 3)(x − 5).

Write the four products: x × x = x², x × (−5) = −5x, 3 × x = 3x and 3 × (−5) = −15.
Add them up: x² − 5x + 3x − 15.
Join the x terms. The numbers in front are −5 and 3, and −5 + 3 = −2.
So the x terms make −2x, and (x + 3)(x − 5) = x² − 2x − 15.

Answer: x² − 2x − 15

You can test an expansion with a number. Put the same value of x into the brackets and into your answer. A right expansion gives the same number both ways.

The first two forms agree. The third gives 23, so it is a wrong expansion.

A bracket squared is the bracket times itself. (2x + 1)² means (2x + 1)(2x + 1). It has four products, not two. The two middle products are equal, and together they make the middle term.

The two hatched middle cells are 2x and 2x. They make 4x.

Worked example

Expand (2x + 1)².

Write the square as two brackets: (2x + 1)(2x + 1).
First times first: 2x × 2x = 4x², because 2 × 2 = 4.
Last times last: 1 × 1 = 1.
The two middle products are 2x × 1 = 2x and 1 × 2x = 2x. Add their numbers: 2 + 2 = 4.
So (2x + 1)² = 4x² + 4x + 1.

Answer: 4x² + 4x + 1

See it three ways

With things

Cut a rectangle that is x + 4 long and x + 2 wide into four parts, and add the parts.

As a picture

In words

Multiply every term in the first bracket by every term in the second. That gives four products. Then join the two like terms, 4x and 2x.

In symbols

(x + 4)(x + 2) = x² + 2x + 4x + 8 = x² + 6x + 8.

Try it

Test an expansion with numbers

You need: paper and a pencil.

Take (x + 3)(x + 2) and its expansion x² + 5x + 6.
Put x = 1 into both. Work out 4 × 3, and then 1 + 5 + 6.
Try x = 2, and then x = 10. Write the three results in a table.

What you will see: The brackets and the expansion give the same number every time. If they ever differ, there is a mistake in the expansion.

If you cannot try it, this is what happens.

A wrong idea many people have

(x + 3)² = x² + 9.

Why it feels right: Squaring one number squares it, so it seems that squaring a bracket squares each part inside.

The test that breaks it: Put x = 1. Then (1 + 3)² = 16, but 1² + 9 = 10. They differ, so x² + 9 is wrong.

What is true: A bracket squared is the bracket times itself, with four products. The two middle terms, 3x and 3x, make 6x. So (x + 3)² = x² + 6x + 9.

A wrong idea many people have

(x + 3)(x − 5) = x² − 2x + 15.

Why it feels right: The last term is 3 times 5, so it feels as if it should be a positive 15.

The test that breaks it: Put x = 4. The brackets give 7 × (−1) = −7. But 4² − 2 × 4 + 15 = 23. The two numbers differ.

What is true: The last product is 3 × (−5), and a positive times a negative is negative. It is −15. Always take the sign with the term.

Key words

expand: to multiply out brackets, so that no brackets are left.
term: one part of an expression that is added or taken away, such as 3x or −15.
like terms: terms with the same letters and powers, such as 3x and −5x. They can be joined.
quadratic expression: an expression whose highest power of x is x², such as x² + 5x + 6.

Back to your guess

The answer is B: x² + 5x + 6: multiply every term by every term, then join the like terms.

Every term in the first bracket multiplies every term in the second. That gives four products: x², 2x, 3x and 6. The two middle terms are alike and join to make 5x. So (x + 3)(x + 2) = x² + 5x + 6.

Keep it

Copy this into your exercise book.

Multiply every term by every term. Keep each sign with its term. Join the like terms. A squared bracket is the bracket times itself, with four products. Check with a number.

Draw: A grid with x and +3 across the top, x and −5 down the side, and four products in the cells.

M4.9.2 Chapter 2

Factorising quadratics

I can expand (x + 3)(x + 4) to get x² + 7x + 12. How do I go backwards?

Your guess

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

Your guess

A Divide every term by x.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Find two numbers that add to 12 and multiply to 7.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Find two numbers that multiply to 12 and add to 7.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Factorise x² + 7x + 12
Factorise a difference of two squares, like x² − 49
Factorise 2x² + 7x + 3

To factorise is to write an expression as a product of brackets. It undoes expanding. The brackets are the factors of the expression.

Look again at the rectangle for x² + 7x + 12. It is x + 4 long and x + 3 wide. The parts hold x², 4x, 3x and 12.

The two numbers 4 and 3 multiply to the last part, 12. They add to the x parts, 4x + 3x = 7x.

This is the key. In x² + 7x + 12, the last number 12 is the product of the two numbers in the brackets. The x number 7 is their sum. So look for two numbers with product 12 and sum 7.

List the pairs with product 12. Only the pair 3 and 4 has sum 7.

Draw it first

From x² + 7x + 12 to two brackets

1 Start with the quadratic, x² + 7x + 12.
2 Find two numbers with product 12 and sum 7. They are 3 and 4.
3 Write each number in a bracket with x: (x + 3)(x + 4).

Worked example

Factorise x² + 7x + 12.

The last number is 12 and the x number is 7. Find two numbers with product 12 and sum 7.
The pairs with product 12 are 1 and 12, 2 and 6, and 3 and 4. Their sums are 13, 8 and 7.
Only the pair 3 and 4 adds up to 7: 3 + 4 = 7.
So x² + 7x + 12 = (x + 3)(x + 4).
Check by expanding: x² + 4x + 3x + 12 = x² + 7x + 12.

Answer: (x + 3)(x + 4)

The signs tell you what kind of numbers to look for. Read the last number first, and then the x number.

When the last number is negative, the two numbers have different signs. The one further from 0 takes the sign of the x number.

Worked example

Factorise x² − 5x + 6.

The last number is +6, so the two numbers have the same sign. The x number is −5, so both are negative.
Try −2 and −3. Their product is (−2) × (−3) = 6.
Their sum is (−2) + (−3) = −5, which is the x number.
So x² − 5x + 6 = (x − 2)(x − 3).
Check by expanding: x² − 3x − 2x + 6 = x² − 5x + 6.

Answer: (x − 2)(x − 3)

Worked example

Factorise x² + 2x − 15.

The last number is −15, so one number is positive and one is negative. The x number is +2, so the positive one is further from 0.
The pairs with product 15 are 1 and 15, and 3 and 5. Make the one further from 0 positive: 15 and −1, or 5 and −3.
Try 5 and −3. Their product is 5 × (−3) = −15.
Their sum is 5 + (−3) = 2, which is the x number.
So x² + 2x − 15 = (x − 3)(x + 5).
Check with x = 1: (1 − 3)(1 + 5) = −12, and 1 + 2 − 15 = −12.

Answer: (x − 3)(x + 5)

Some quadratics have no x term, like x² − 49. The number 49 is a square, since 7 × 7 = 49. Try the brackets (x + 7)(x − 7) in a grid.

The hatched cells are +7x and −7x. They cancel, so only x² and −49 are left.

This is a difference of two squares: one square minus another. It always factorises the same way. a² − b² = (a + b)(a − b). The middle terms cancel because the signs are opposite.

Worked example

Factorise x² − 49.

The number 49 is a square, so x² − 49 is a difference of two squares. The square root of 49 is 7.
Use two brackets with +7 in one and −7 in the other: (x + 7)(x − 7).
Check by expanding: x² − 7x + 7x − 49 = x² − 49. The middle terms cancel.

Answer: (x + 7)(x − 7)

It works only when there is a minus between two squares. The expression x² + 49 has no brackets of this kind, because no real numbers multiply to 49 and add to 0.

When the number in front of x² is not 1, a grid helps. The quadratic 2x² + 7x + 3 comes from the brackets (2x + 1)(x + 3). Here is the grid.

The two hatched cells, 6x and x, add to the middle term 7x. They multiply to 6x², which is 2x² × 3.

Worked example

Factorise 2x² + 7x + 3.

Multiply the first number and the last number: 2 × 3 = 6.
Find two numbers with product 6 and sum 7. Try 1 and 6: their product is 6 and their sum is 1 + 6 = 7.
Split the middle term: 2x² + x + 6x + 3.
Take out a common factor from each pair: x(2x + 1) + 3(2x + 1).
Both parts have the factor (2x + 1), so the answer is (2x + 1)(x + 3).
Check with x = 1: (2 + 1)(1 + 3) = 12, and 2 + 7 + 3 = 12.

Answer: (2x + 1)(x + 3)

Always look for a common factor first. In 3x² + 12x + 9 every term divides by 3. So 3x² + 12x + 9 = 3(x² + 4x + 3) = 3(x + 1)(x + 3).

See it three ways

With things

Lay out tiles: one big square x², eight strips of x, and 15 small squares. Arrange them into one rectangle.

As a picture

In words

The last number 15 is the product of the two numbers 3 and 5. The x number 8 is their sum. So the brackets are x + 3 and x + 5.

In symbols

x² + 8x + 15 = (x + 3)(x + 5).

Try it

A difference of two squares at work

You need: paper and a pencil.

Work out 51 × 49 on paper. It takes a while.
Now think of x = 50. Then 51 × 49 is (x + 1)(x − 1), which is x² − 1.
Work out 50 × 50 − 1. Compare the two answers.

What you will see: Both ways give 2 499. A difference of two squares turns a hard product into a square minus a small number.

If you cannot try it, this is what happens.

A wrong idea many people have

x² − 49 = (x − 7)(x − 7).

Why it feels right: The number 7 times 7 is 49, so it seems that two matching brackets should work.

The test that breaks it: Put x = 10. Then x² − 49 = 51, but (10 − 7)(10 − 7) = 9. The two numbers differ.

What is true: The two brackets need opposite signs, so that the middle terms cancel. The answer is (x + 7)(x − 7).

A wrong idea many people have

x² − 5x + 6 = (x + 2)(x + 3), because 2 × 3 = 6 and 2 + 3 = 5.

Why it feels right: The numbers 2 and 3 do multiply to 6 and add to 5, so they look right.

The test that breaks it: Expand it: (x + 2)(x + 3) = x² + 5x + 6. The middle term is +5x, but the quadratic has −5x.

What is true: The x number is −5, so the two numbers must add to −5. Use −2 and −3. Always expand to check the signs.

Key words

factorise: to write an expression as a product of brackets.
factor: a bracket or number that divides an expression exactly.
difference of two squares: an expression like x² − 49: one square minus another square.
common factor: a factor that every term has, such as 3 in 3x² + 12x + 9.

Back to your guess

The answer is C: Find two numbers that multiply to 12 and add to 7.

In (x + 3)(x + 4) the last numbers multiply: 3 × 4 = 12. The x terms add: 3x + 4x = 7x. So going backwards, find two numbers that multiply to 12 and add to 7. They are 3 and 4.

Keep it

Copy this into your exercise book.

Find two numbers that multiply to the last number and add to the x number. Check the signs. Look for a common factor first. Expand to check.

Draw: The table of pairs for 12 with the pair 3 and 4 marked, and the brackets (x + 3)(x + 4).

M4.9.3 Chapter 3

Using factorising

I factorised x² − 5x + 6 = 0 into (x − 2)(x − 3) = 0. What does that say about x?

Your guess

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

Your guess

A x = 2 or x = 3.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x = −2 or x = −3.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x = 2 and x = 3 at the same time.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve x² − 5x + 6 = 0 by factorising
Explain why ab = 0 means a = 0 or b = 0
Simplify (x² − 9)/(x + 3) by factorising

Factorising gives a way to solve equations. Take x² − 5x + 6 = 0. The left side factorises to (x − 2)(x − 3). So the equation says that two numbers multiply to give 0.

Try some products. If neither number is 0, the product is not 0. The product is 0 only when at least one of the numbers is 0.

The product is 0 in the two rows where a number is 0, and in no other row.

Here is why. Say a × b = 0 and a is not 0. Divide both sides by a. That gives b = 0. So if a is not 0, then b must be. At least one of a and b is 0.

Draw it first

Solve x² − 5x + 6 = 0

1 Start with the equation. It already has 0 on the right.
2 Factorise the left side: (x − 2)(x − 3) = 0.
3 A product is 0 only if a factor is 0. So x − 2 = 0 or x − 3 = 0.
4 Solve each one. x = 2 or x = 3. These are the roots of the equation.

Worked example

Solve x² − 5x + 6 = 0.

Factorise the left side: (x − 2)(x − 3) = 0.
A product is 0 only when one factor is 0. So x − 2 = 0 or x − 3 = 0.
Solve each one: x = 2 or x = 3.
Check x = 3 in the equation: 3 × 3 − 5 × 3 + 6 = 9 − 15 + 6 = 0.

Answer: x = 2 or x = 3

You can see the roots on a graph. The curve y = x² − 5x + 6 meets the x axis where y = 0. That is exactly where the equation is true.

The curve meets the x axis at A, where x = 2, and at B, where x = 3. These are the roots.

The method needs 0 on one side. If it is not there, make it. Take everything to the left, so that the right side is 0, and then factorise.

Worked example

Solve x² + 12 = 7x.

Take 7x from both sides, so that one side is 0: x² − 7x + 12 = 0.
Two numbers with product 12 and sum −7 are −3 and −4. So (x − 3)(x − 4) = 0.
Put each factor equal to 0 and solve: x − 3 = 0 gives x = 3, and x − 4 = 0 gives x = 4.
Check x = 4 in the first equation: 4 × 4 + 12 = 28, and 7 × 4 = 28.

Answer: x = 3 or x = 4

See it three ways

With things

Try the whole numbers 1 to 4 in (x − 1)(x − 4). Write down what each gives.

As a picture

In words

The product is 0 for x = 1 and for x = 4. It is 0 only when x − 1 = 0 or x − 4 = 0.

In symbols

(x − 1)(x − 4) = 0 gives x − 1 = 0 or x − 4 = 0, so x = 1 or x = 4.

A difference of two squares gives two answers with opposite signs. Take x² − 16 = 0. It factorises as (x + 4)(x − 4) = 0.

Worked example

Solve x² − 16 = 0.

Write 16 as 4 × 4. The left side is a difference of two squares: (x + 4)(x − 4) = 0.
Put each factor equal to 0: x + 4 = 0 or x − 4 = 0.
So x = −4 or x = 4. Both work, because (−4) × (−4) = 16 and 4 × 4 = 16.

Answer: x = −4 or x = 4

Factorising also simplifies fractions made of expressions. Take (x² − 9)/(x + 3). The top is a difference of two squares, so it factorises. Then a bracket that is on the top and the bottom cancels.

The bracket (x + 3) is a factor of the whole top and of the whole bottom, so it cancels.

Worked example

Simplify (x² − 9)/(x + 3).

Factorise the top: x² − 9 = (x + 3)(x − 3).
Write the fraction with the top factorised: (x + 3)(x − 3) over (x + 3).
Cancel the bracket (x + 3) on the top and on the bottom.
What is left is x − 3. This holds when x is not −3, because the bottom cannot be 0.
Check with x = 5: (25 − 9)/(5 + 3) = 16/8 = 2, and 5 − 3 = 2.

Answer: x − 3

You can only cancel a whole bracket or a whole number that multiplies all of the top and all of the bottom. A term that is added cannot be cancelled.

Try it

Find where a quadratic is 0

You need: paper and a pencil.

Work out x² − 5x + 6 for x = 1, 2, 3 and 4.
Write each result in a table.
Look for the values of x that give 0.

What you will see: The results are 2, 0, 0 and 2. The zeros are at x = 2 and x = 3, the roots you found by factorising.

If you cannot try it, this is what happens.

A wrong idea many people have

To solve x² = 5x, divide both sides by x. So x = 5.

Why it feels right: Dividing both sides by the same thing keeps an equation true, and it makes the equation shorter.

The test that breaks it: Put x = 0. Then x² = 0 and 5x = 0, so x = 0 is also an answer. Dividing by x lost it.

What is true: Do not divide both sides by x: x may be 0, and that answer is lost. Move everything to one side: x² − 5x = 0. Take out x: x(x − 5) = 0. So x = 0 or x = 5.

A wrong idea many people have

(x − 1)(x − 2) = 6 gives x − 1 = 6 or x − 2 = 6.

Why it feels right: It copies the method for 0. Each bracket seems to be allowed to take the 6.

The test that breaks it: Take x = 7 from x − 1 = 6. Then (7 − 1)(7 − 2) = 6 × 5 = 30, not 6.

What is true: Splitting into brackets works only when the product is 0. Expand and make one side 0 first: x² − 3x − 4 = 0. Then (x − 4)(x + 1) = 0.

A wrong idea many people have

(x + 3)/3 = x.

Why it feels right: There is a 3 on the top and a 3 on the bottom, so they seem to cancel.

The test that breaks it: Put x = 3. Then (3 + 3)/3 = 2, but the answer x gives 3.

What is true: The 3 on the top is added to x. It does not multiply the whole top, so it cannot cancel. Only a factor of the whole top can cancel.

Key words

roots: the solutions of an equation like x² − 5x + 6 = 0: the values of x that make it true.
cancel: to divide the top and the bottom of a fraction by the same factor.

Back to your guess

The answer is A: x = 2 or x = 3.

Two numbers multiply to 0 only if one of them is 0. So x − 2 = 0 or x − 3 = 0, which gives x = 2 or x = 3. Both answers work, but not at the same time.

Keep it

Copy this into your exercise book.

Make one side 0. Factorise. A product is 0 only if a factor is 0, so put each bracket equal to 0. Do not divide both sides by x: x may be 0. Cancel only whole brackets.

Draw: The equation x² − 5x + 6 = 0, the brackets (x − 2)(x − 3) = 0, and the two answers.

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

A (x + 3)² + 2
You picked A. Read on. At the end, the chapter comes back to your guess.
B (x + 6)² − 34
You picked B. Read on. At the end, the chapter comes back to your guess.
C (x + 3)² − 7
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 A square x by x, and two strips 3 by x, make x² + 3x + 3x = x² + 6x. A corner is missing.
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.

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

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.

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

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.

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

Write x² + 6x + 2 as (x + 3)² − 7. The equation is (x + 3)² − 7 = 0.
Add 7 to both sides: (x + 3)² = 7.
Take the square root of both sides: x + 3 = √7 or x + 3 = −√7.
Use √7 ≈ 2.6458. Then x ≈ 2.6458 − 3 = −0.35, or x ≈ −2.6458 − 3 = −5.65.
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

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.

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

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.

M4.9.5 Chapter 5

The quadratic formula

x² + 4x + 1 = 0 does not factorise with whole numbers. Can I still solve it every time?

Your guess

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

Your guess

A Yes. One formula uses the three numbers in the equation.
You picked A. Read on. At the end, the chapter comes back to your guess.
B No. Only quadratics that factorise can be solved.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes. Try x = 1, 2, 3 and so on until one works.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve any quadratic equation with the formula
Use b² − 4ac to say how many solutions there are
Form and solve a quadratic from an area problem

Some quadratics do not factorise with whole numbers. Completing the square always works, but it takes many steps each time. The quadratic formula does all of that work once, for every quadratic.

Write any quadratic as ax² + bx + c = 0. The number a is in front of x², the number b is in front of x, and c stands alone. These are the coefficients. Read each one with its sign.

A letter with no number in front, like x, has the number 1. A minus sign belongs to the number after it.

The solutions of ax² + bx + c = 0 are x = (−b ± √(b² − 4ac)) ÷ (2a). The sign ± means two answers: one with a plus and one with a minus. The whole top is divided by 2a. Here is why the formula is true.

Draw it first

Where the formula comes from, part one

1 Start with any quadratic, ax² + bx + c = 0, where a is not 0.
2 Multiply both sides by 4a. Every term is multiplied, so the equation becomes 4a²x² + 4abx + 4ac = 0.
3 Add b² to both sides, then take away 4ac from both sides. The left side is now 4a²x² + 4abx + b², and the right side is b² − 4ac.
4 The left side is a perfect square. Check it with a grid as in chapter 1: (2ax + b)(2ax + b) = 4a²x² + 2abx + 2abx + b². So (2ax + b)² = b² − 4ac.

Now take the square root of both sides. This works only when b² − 4ac is 0 or more, because no real number squared is below 0. A positive number has two square roots, so write ±.

Draw it first

Where the formula comes from, part two

1 We have (2ax + b)² = b² − 4ac.
2 Take the square root of both sides, which needs b² − 4ac to be 0 or more. There are two square roots, so write ±.
3 Take b from both sides and divide by 2a. That is the formula.

To use the formula, first read off a, b and c. Then work out the part under the root, b² − 4ac, on its own. Last, put everything in.

Worked example

Solve x² + 3x − 10 = 0 with the formula.

Read off the numbers: a = 1, b = 3 and c = −10.
Work out b² − 4ac = 3 × 3 − 4 × 1 × (−10) = 9 + 40 = 49.
The square root of 49 is 7.
Use the formula: x = (−3 + 7)/2 or x = (−3 − 7)/2.
So x = 4/2 = 2 or x = −10/2 = −5.
Check with x = 2: 2 × 2 + 3 × 2 − 10 = 4 + 6 − 10 = 0.

Answer: x = 2 or x = −5

This quadratic also factorises: (x + 5)(x − 2) = 0 gives the same answers. The formula also works when factorising does not, and it is quicker than completing the square.

Take care when b is negative. The formula starts with −b, and −(−4) is +4. The part b² is the square of b, so (−4)² = 16, which is positive.

Worked example

Solve x² − 4x − 3 = 0. Give the answers to 2 decimal places.

Read off the numbers: a = 1, b = −4 and c = −3.
Work out b² − 4ac = (−4) × (−4) − 4 × 1 × (−3) = 16 + 12 = 28.
The square root of 28 is about 5.2915.
Use the formula. Since b = −4, −b = 4. So x = (4 + 5.2915)/2 or x = (4 − 5.2915)/2.
Work out each one: x = 9.2915/2 ≈ 4.65, or x = −1.2915/2 ≈ −0.65.

Answer: x ≈ 4.65 or x ≈ −0.65

In every example so far a = 1, so the bottom of the formula was 2. When a is not 1, the bottom is 2a, which changes. Here is one with a = 2.

Worked example

Solve 2x² + 3x − 2 = 0 with the formula.

Read off the numbers: a = 2, b = 3 and c = −2.
Work out b² − 4ac = 9 − 4 × 2 × (−2) = 9 + 16 = 25.
The square root of 25 is 5.
The bottom of the formula is 2a, and 2a = 2 × 2 = 4.
So x = (−3 + 5)/4 or x = (−3 − 5)/4.
Work out each one: x = 2/4 = 0.5 or x = −8/4 = −2.
Check with x = 0.5: 2 × 0.25 + 1.5 − 2 = 0.

Answer: x = 0.5 or x = −2

The part under the root is called the discriminant, D = b² − 4ac. It tells you how many solutions there are before you finish. Take three quadratics that differ only in the last number.

D above 0 gives two solutions. D equal to 0 gives one. D below 0 gives none.

Here are their graphs. A solution is a place where the curve meets the x axis. First D = 4. The square root of 4 is 2, so the formula gives two different answers.

D = 4. The curve crosses the x axis twice, at x = 1 and x = 3.

When D = 0, the square root is 0, so plus and minus give the same answer. The curve just touches the x axis.

D = 0. The curve touches the x axis once, at x = 2.

When D is below 0, the formula needs the square root of a negative number. No real number squared makes a negative, so there is no real solution. The curve stays above the x axis.

D = −4. The curve never meets the x axis, so the equation has no solution.

Worked example

How many solutions does 2x² + 3x + 5 = 0 have?

Read off the numbers: a = 2, b = 3 and c = 5.
Work out b² − 4ac = 3 × 3 − 4 × 2 × 5 = 9 − 40 = −31.
The discriminant is below 0, so the formula needs the square root of a negative number. There is no such real number, so there is no real solution.

Answer: No real solution. The discriminant is −31.

Quadratic equations come from area problems. A rectangle is 3 m longer than it is wide, and its area is 28 m². Call the width x metres. Then the length is x + 3 metres.

Area is length times width, so x(x + 3) = 28.

Worked example

A rectangle is 3 m longer than it is wide. Its area is 28 m². How long and wide is it?

Let x be the width in metres. The length is x + 3 metres.
Area = length × width, so x(x + 3) = 28.
Expand and make one side 0: x² + 3x − 28 = 0.
Use the formula with a = 1, b = 3 and c = −28. Then b² − 4ac = 9 + 112 = 121, and the square root of 121 is 11.
So x = (−3 + 11)/2 = 4, or x = (−3 − 11)/2 = −7.
A width cannot be negative, so x = 4. The rectangle is 4 m wide and 7 m long.
Check the area: 4 × 7 = 28.

Answer: The rectangle is 4 m wide and 7 m long.

See it three ways

With things

Draw the curve y = x² − 6x + 5 from a table of values. Read where it crosses the x axis.

As a picture

In words

The curve crosses the x axis at 1 and 5, so those are the solutions. The formula gives the same: b² − 4ac = 36 − 20 = 16, and the square root of 16 is 4.

In symbols

x = (6 ± 4)/2, so x = 5 or x = 1.

Try it

Solve x² + 4x + 1 = 0 and test it

You need: paper, a pencil and a calculator.

Read off a = 1, b = 4 and c = 1. Work out b² − 4ac = 16 − 4 = 12.
Use the formula with √12 ≈ 3.4641: x = (−4 + 3.4641)/2 ≈ −0.27, or x = (−4 − 3.4641)/2 ≈ −3.73.
Put each answer back into x² + 4x + 1 and write what you get.

What you will see: Both answers give a value very close to 0. They are not exactly 0 only because the answers are rounded.

If you cannot try it, this is what happens.

A wrong idea many people have

When b = −6, b² = −36.

Why it feels right: The minus sign seems to stay in front of the number, so the square looks negative too.

The test that breaks it: Work out (−6) × (−6). A negative times a negative is positive, so it is 36.

What is true: The square of a negative number is positive. In the formula, b² is the square of the whole of b, sign included. With b = −6, b² = 36.

A wrong idea many people have

x = −b ± √(b² − 4ac) / 2a: only the root is divided by 2a.

Why it feels right: Written on one line, the division seems to belong to the last part only.

The test that breaks it: Take x² − 2x − 8 = 0. Dividing only the root gives 2 ± 6/2 = 2 ± 3, so x = 5. But 5 × 5 − 2 × 5 − 8 = 7, not 0.

What is true: The whole top, −b ± √(b² − 4ac), is divided by 2a. For this equation x = (2 ± 6)/2, so x = 4 or x = −2.

A wrong idea many people have

A quadratic always has two solutions, because of the ±.

Why it feels right: The formula has a plus or minus, so it seems that every quadratic gives two answers.

The test that breaks it: Try x² + 1 = 0. This needs x² = −1, but no real number squared makes −1. The formula has D = −4, with no real square root.

What is true: The ± gives two answers only when the discriminant is above 0. When D = 0 there is one answer, and when D is below 0 there is no real solution.

Key words

coefficient: the number in front of a letter, such as the 5 in 5x, with its sign.
discriminant: the number b² − 4ac, which tells you how many solutions a quadratic has.

Back to your guess

The answer is A: Yes. One formula uses the three numbers in the equation.

One formula solves every quadratic. It comes from completing the square on the letters a, b and c, so it works for any numbers. Trying whole numbers is slow, and it misses answers that are not whole.

Keep it

Copy this into your exercise book.

Read off a, b and c with their signs. Work out b² − 4ac first. Then x = (−b ± √(b² − 4ac)) ÷ (2a). Above 0: two answers. At 0: one. Below 0: no real answer.

Draw: Three curves: one crossing the x axis twice, one touching it, one above it.

M4.9.6 Chapter 6

Graphs of quadratics

I plot the points of y = x² − 2x − 3 on a grid. What shape do they make?

Your guess

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

Your guess

A A straight line, like y = 2x + 1.
You picked A. Read on. At the end, the chapter comes back to your guess.
B A smooth curve with a lowest point, shaped like a U.
You picked B. Read on. At the end, the chapter comes back to your guess.
C A V with a sharp point at the bottom.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Draw y = x² − 2x − 3 from a table of values
Find the roots, turning point and y axis crossing from a graph

A rule like y = x² − 2x − 3 gives a y for every x. To draw its graph, make a table of values. Work out y for several values of x, then plot each pair as a point.

The table for y = x² − 2x − 3. Look at how the y values fall, then rise again.

Be careful with negative x. When x = −2, x² is (−2) × (−2) = 4, which is positive. And −2x is −2 × (−2) = +4. So y = 4 + 4 − 3 = 5.

Draw it first

From the table to the curve

1 Plot each pair from the table as a point. Go across first, then up or down.
2 Join the points with one smooth curve. Never use a ruler, because the graph bends between the points.

The curve of a quadratic is called a parabola. It has a lowest point, or a highest point if the x² term is negative. The curve is a mirror image on the two sides of that point.

Worked example

Make a table for y = x² − 2x − 3 and find the lowest point of the curve.

Work out y when x = −2: (−2) × (−2) − 2 × (−2) − 3 = 4 + 4 − 3 = 5.
Work out y when x = 1: 1 × 1 − 2 × 1 − 3 = 1 − 2 − 3 = −4.
The other values are in the table: y = 0 when x = −1 and x = 3, and y = 5 when x = 4.
The table is symmetric. Look: y = −3 at x = 0 and at x = 2. Also y = 0 at x = −1 and at x = 3. So the turning point is halfway, at x = 1, where y = −4.
So the lowest point is (1, −4).

Answer: The lowest point is (1, −4).

Now read the graph. Look at where the curve meets the axes, and at its lowest point. Each of these tells you something about the quadratic.

See it three ways

With things

Find the points of the curve y = x² − 2x − 3 where it meets the x axis, meets the y axis, and turns.

As a picture

In words

A and B are the roots, where y = 0: x = −1 and x = 3. C is where the curve crosses the y axis: (0, −3). T is the turning point (1, −4), halfway between the roots. The dashed line x = 1 is the line of symmetry.

In symbols

x² − 2x − 3 = (x + 1)(x − 3), so y = 0 at x = −1 and x = 3. At x = 0, y = −3. And x² − 2x − 3 = (x − 1)² − 4, so the lowest point is (1, −4).

So each part of the graph links to work you have done. The roots come from factorising. The y axis crossing is the last number, c. The turning point comes from completing the square.

The turning point is halfway between the roots, because of the mirror symmetry. Halfway between x = −1 and x = 3 is (−1 + 3)/2 = 1. This is a quick way to find x for the turning point. The lowest table value is not always the turning point. Find x by symmetry first.

If the x² term is negative, the curve is upside down. Take y = −x² + 2x + 3. It has a highest point, or maximum, instead of a lowest point.

The curve of y = −x² + 2x + 3. With −x², it opens downwards. T is a highest point, at (1, 4).

Try it

Draw your own parabola

You need: squared paper, or paper and a ruler, a pencil.

Make a table for y = x² − 4, with x from −3 to 3.
Plot the seven points. Join them with a smooth curve.
Look at the table and the curve. Where are the roots, and where is the lowest point?

What you will see: The y values read the same going up and down the table from the middle. The roots are x = −2 and x = 2. The lowest point is (0, −4).

If you cannot try it, this is what happens.

A wrong idea many people have

Join the points of a quadratic with a ruler.

Why it feels right: A ruler gives neat lines, and the points are exact, so it seems that joining them is safe.

The test that breaks it: Take x = 0.5. The curve has y = 0.25 − 1 − 3 = −3.75. A ruler line from (0, −3) to (1, −4) gives −3.5 there. They differ.

What is true: A parabola bends between the points. Join them with one smooth curve, and add more points where it turns.

A wrong idea many people have

The lowest point of the curve is where it crosses the y axis.

Why it feels right: The y axis crossing is the one point that is quick to find, so it seems to be the important one.

The test that breaks it: The curve y = x² − 2x − 3 crosses the y axis at (0, −3). But the point (1, −4) is lower.

What is true: The crossing of the y axis is just the point where x = 0. The lowest point is where the curve turns, halfway between the roots.

Key words

parabola: the smooth curve of a quadratic, shaped like a U or an upside down U.
turning point: the lowest point of a U shaped curve, or the highest point of an upside down one.
line of symmetry: a line that cuts a shape into two mirror images, such as x = 1 for this curve.

Back to your guess

The answer is B: A smooth curve with a lowest point, shaped like a U.

The x² makes the graph bend. Its points lie on a smooth curve, shaped like a U, called a parabola. The curve is lowest at the point (1, −4), and there it turns, with no sharp point.

Keep it

Copy this into your exercise book.

Make a table of values, plot the points, join them with a smooth curve. Roots: where y = 0. Crossing of the y axis: the last number. Turning point: halfway between the roots.

Draw: A U shaped curve with its two roots, its y axis crossing, and its turning point marked and named.

M4.9.7 Chapter 7

Quadratic sequences

The sequence 2, 6, 12, 20 goes up by more each time. Is there a rule for the nth term?

Your guess

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

Your guess

A No. Only sequences that go up by the same amount have a rule.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes, but only by guessing.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes. The differences of the differences are all the same and not 0, and that gives the rule.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use second differences to spot a quadratic sequence
Find the nth term of a sequence like 2, 6, 12, 20

In an earlier book, a sequence like 3, 7, 11, 15 went up by 4 each time. Its first differences are all 4. That kind of sequence has an nth term like 4n − 1.

The first differences are all +4. This sequence goes up in equal steps.

Its second differences are all 0, because the steps never change.

Now take 2, 6, 12, 20. It goes up by 4, then 6, then 8. The differences are not equal. Look at the differences of the differences. They are the second differences.

Draw it first

First and second differences

1 Write the terms of the sequence in a row.
2 Subtract each term from the next one. These are the first differences: 4, 6 and 8.
3 Do it again with the first differences. The second differences are 2 and 2. They are the same.

When the second differences are all the same, and not 0, the sequence is a quadratic sequence. Its nth term has an n² in it. Why? Look at the square numbers, 1, 4, 9, 16, 25.

The second differences of n² are all +2.

For n², the second difference is always 2. The first differences of n² are 3, 5, 7, 9, the odd numbers, and they go up by 2 each time. For 3n² it would be 3 × 2 = 6. For a n², the second difference is 2a. A part like bn + c goes up by the same b each step, so it adds nothing to the second differences. So the n² number is half the second difference.

Not every sequence has constant second differences. Take the cubes, 1, 8, 27, 64, 125. Their second differences change, so they are not quadratic.

The second differences, 12, 18 and 24, are not all the same. So the cubes are not a quadratic sequence.

Worked example

Find the nth term of 2, 6, 12, 20.

The first differences are 4, 6, 8. The second differences are 2, 2. They are the same, so the sequence is quadratic.
The second difference is 2, so the n² number is half of it: 2 ÷ 2 = 1. The nth term starts with n².
The values of n² for n = 1, 2, 3, 4 are 1, 4, 9, 16.
Take these from the sequence: 2 − 1 = 1, 6 − 4 = 2, 12 − 9 = 3 and 20 − 16 = 4.
The results 1, 2, 3, 4 are the numbers n. So the nth term is n² + n.
Check with n = 4: 16 + 4 = 20.

Answer: n² + n

Worked example

Find the nth term of 3, 8, 15, 24.

The first differences are 5, 7, 9. The second differences are 2, 2. The sequence is quadratic.
Start with n². Its number is half of the second difference: 2 ÷ 2 = 1.
The values of n² are 1, 4, 9, 16.
Take them from the sequence: 3 − 1 = 2, 8 − 4 = 4, 15 − 9 = 6 and 24 − 16 = 8.
The results 2, 4, 6, 8 go up in 2s from 2. That is 2n.
So the nth term is n² + 2n.
Check with n = 3: 9 + 6 = 15.

Answer: n² + 2n

When the second difference is not 2, the n² number changes. A second difference of 4 means 4 ÷ 2 = 2, so the nth term starts with 2n².

Worked example

Find the nth term of 3, 10, 21, 36.

The first differences are 7, 11, 15. The second differences are 4, 4. The sequence is quadratic.
Start with 2n². Its number is half of the second difference: 4 ÷ 2 = 2.
The values of 2n² are 2, 8, 18, 32.
Take them from the sequence: 3 − 2 = 1, 10 − 8 = 2, 21 − 18 = 3 and 36 − 32 = 4.
The results 1, 2, 3, 4 are the numbers n.
So the nth term is 2n² + n.
Check with n = 4: 2 × 16 + 4 = 36.

Answer: 2n² + n

See it three ways

With things

Write the sequence 2, 6, 12, 20, and find the next term by adding on the differences.

As a picture

In words

The differences go up by 2 each time: 4, 6, 8, 10. So the next difference is 10, and the next term is 20 + 10 = 30.

In symbols

The nth term is n² + n. For n = 5, 5² + 5 = 25 + 5 = 30.

Try it

Make a quadratic sequence

You need: paper and a pencil.

Take the rule n² + 3n. Work it out for n = 1, 2, 3, 4 and 5.
Write the five terms in a row. Find the first differences, then the second differences.
Compare the second difference with the n² number in the rule.

What you will see: The terms are 4, 10, 18, 28, 40. The second differences are all 2, which is twice the number in front of n². The rule and the differences agree.

If you cannot try it, this is what happens.

A wrong idea many people have

The second difference is the number in front of n². For 2, 6, 12, 20 the nth term starts with 2n².

Why it feels right: The second difference, 2, stands out in the table, so it seems to be the number in front of n².

The test that breaks it: Put n = 2 into 2n². It gives 2 × 4 = 8, but the second term of the sequence is 6. So 2n² is too big.

What is true: The sequence n² has second difference 2. So a n² has second difference 2a. Halve the second difference to find a. Here a = 1.

A wrong idea many people have

If the first differences are not all the same, the sequence has no rule.

Why it feels right: For 3, 7, 11, 15 the first differences are all the same, so a sequence where they change looks like a mess.

The test that breaks it: The sequence 2, 6, 12, 20 has first differences 4, 6, 8, which change. But its second differences are 2, 2, and the rule n² + n gives every term.

What is true: Take differences again. When the second differences are all the same and not 0, there is a quadratic rule.

Key words

first differences: what you get by taking each term of a sequence from the next one.
second differences: the differences of the first differences.
quadratic sequence: a sequence whose second differences are all the same and not 0, so that its nth term has an n².

Back to your guess

The answer is C: Yes. The differences of the differences are all the same and not 0, and that gives the rule.

The differences of 2, 6, 12, 20 are 4, 6, 8. Take differences again and you get 2, 2. When these second differences are all the same and not 0, the rule is a quadratic. There is a method to find it. Here the nth term is n² + n.

Keep it

Copy this into your exercise book.

Find the differences twice. If the second differences match and are not 0, it is quadratic. Halve the second difference for the n² number. Take n² away, then find the rest.

Draw: The terms 2, 6, 12, 20 with first differences 4, 6, 8 and second differences 2, 2.

M4.9.8 Chapter 8

A line and a curve

The line y = x + 2 and the curve y = x² cross. How do I find where?

Your guess

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

Your guess

A Find the x where x² = x + 2, then find y from the line.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Add the two equations together.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Draw the graph and read the answer. There is no other way.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve a linear and a quadratic equation together
Solve a quadratic inequality like x² < 9

A point where a line meets a curve is on both. So its x and its y fit both equations. For y = x² and y = x + 2, the two values of y must be equal.

The line cuts the curve at P and Q. We find them with algebra, and the graph checks the answer.

Put the two expressions for y equal. That gives x² = x + 2. It is a quadratic equation. Solve it, and each solution is the x of a meeting point.

Draw it first

Where the line meets the curve

1 Both equations give the same y at a meeting point, so put them equal: x² = x + 2.
2 Move every term to the left, so that the right side is 0.
3 Factorise the left side: (x − 2)(x + 1) = 0.
4 Solve each bracket. x = 2 or x = −1. These are the x values of the two meeting points.

The x values are only half the answer. A meeting point needs a y as well. Put each x into the line, which is the easier equation, to find it.

Worked example

Find where y = x² and y = x + 2 meet.

At a meeting point both equations give the same y. So x² = x + 2.
Move every term to the left: x² − x − 2 = 0.
Factorise: (x − 2)(x + 1) = 0. So x = 2 or x = −1.
Find y from the line y = x + 2. For x = 2, y = 4. For x = −1, y = 1.
Check in the curve y = x². For x = 2, 2 × 2 = 4. For x = −1, (−1) × (−1) = 1.
The curve and the line meet at (2, 4) and (−1, 1).

Answer: (2, 4) and (−1, 1)

Worked example

Find where y = x² − 2x and y = x + 4 meet.

Put the two values of y equal: x² − 2x = x + 4.
Move every term to the left: x² − 3x − 4 = 0.
Factorise: (x − 4)(x + 1) = 0. So x = 4 or x = −1.
Find y from the line y = x + 4. For x = 4, y = 8. For x = −1, y = 3.
Check in the curve y = x² − 2x. For x = 4, 16 − 8 = 8. For x = −1, 1 + 2 = 3.

Answer: (4, 8) and (−1, 3)

A line does not always cut a curve twice. The quadratic you get can have two solutions, one, or none. This links to the discriminant. First, a line with one solution.

Here x² = 2x − 1 gives x² − 2x + 1 = 0, which is (x − 1)² = 0. There is one solution, x = 1. The line touches the curve at P.

A line that touches a curve at one point, and does not cross it, is a tangent. Now a line that misses the curve. Take y = x − 3.

Here x² = x − 3 gives x² − x + 3 = 0. Its discriminant is 1 − 12 = −11, which is below 0. There is no solution, so the line misses the curve.

Now a different question. When is x² less than 9? Draw the curve y = x² and the line y = 9. The curve is below the line between the points where they meet.

The curve is below the line y = 9 between P and Q, which are at x = −3 and x = 3.

See it three ways

With things

Try the whole numbers from −4 to 4 in x². Mark the ones where x² is less than 9.

As a picture

In words

The whole numbers −2, −1, 0, 1 and 2 work. The numbers −3 and 3 give exactly 9, which is not less than 9. So x is between −3 and 3, and not equal to either.

In symbols

x² < 9 means −3 < x < 3.

For x² > 9 it is the other way round. The curve is above the line outside the meeting points. So x is less than −3, or x is more than 3.

x² > 9 means x < −3 or x > 3. The two parts are separate.

Other quadratic inequalities use the same idea. Solve the equation to find the roots. When the x² term is positive, picture the U shaped curve, and read where it is below or above the x axis.

Between A and B the curve is below the x axis, so x² − x − 6 < 0 there.

Worked example

Solve x² − x − 6 < 0.

Solve the equation x² − x − 6 = 0 first: (x − 3)(x + 2) = 0, so x = 3 or x = −2.
These are the roots, where the curve meets the x axis. The curve is a U shape, so it is below the axis between the roots.
So x² − x − 6 is below 0 when x is between −2 and 3: −2 < x < 3.
Check x = 0, which is in the range. Then x² − x − 6 = −6, which is below zero.
Check x = 4, which is outside the range: 16 − 4 − 6 = 6.

Answer: −2 < x < 3

Try it

Test numbers in an inequality

You need: paper and a pencil.

Take x² < 9. Try x = −4, −2, 0, 2 and 4.
For each one work out x², and write yes if it is less than 9 and no if it is not.
Look at which numbers say yes. Are they in one run?

What you will see: The yes answers are −2, 0 and 2. They are in one run, between −3 and 3. The no answers are on the outside.

If you cannot try it, this is what happens.

A wrong idea many people have

x² < 9 means x < 3.

Why it feels right: The square root of 9 is 3, so it seems that x must be below 3.

The test that breaks it: Put x = −5. Then x < 3 is true, but x² = 25, which is not less than 9.

What is true: x can be negative too. The numbers between −3 and 3 all work, so x² < 9 means −3 < x < 3.

A wrong idea many people have

x² > 9 means x > 3.

Why it feels right: It is the same idea as before, with the sign turned round, so only the number 3 seems to matter.

The test that breaks it: Put x = −4. Then x² = 16, which is more than 9. But −4 is not more than 3.

What is true: Large negative numbers also have large squares. So x² > 9 means x < −3 or x > 3.

A wrong idea many people have

When a line meets a curve at x = −1, the meeting point is (−1, 0).

Why it feels right: Solving the quadratic gives only x values, and writing 0 for y seems to complete the point.

The test that breaks it: For y = x + 2 and x = −1, the line gives y = 1. The point (−1, 0) is not on the line.

What is true: Find y from the line, the easier equation. Give each meeting point as a pair, (x, y). Here it is (−1, 1).

Key words

tangent: a line that touches a curve at one point and does not cross it there.

Back to your guess

The answer is A: Find the x where x² = x + 2, then find y from the line.

At a point where they cross, both equations give the same y. So x² = x + 2. That is a quadratic, which you can solve. Then put each x into the line to find y. The crossing points are (−1, 1) and (2, 4).

Keep it

Copy this into your exercise book.

Put the two y values equal, solve, then find y from the line. For x² < 9, x is between −3 and 3. For x² > 9, x is below −3 or above 3.

Draw: A U shaped curve with a flat line across it, and the two meeting points marked and named.

Multiply every pair of terms

Test an expansion with numbers

From x² + 7x + 12 to two brackets

A difference of two squares at work

Solve x² − 5x + 6 = 0

Find where a quadratic is 0

Complete the square

Test the new form with numbers

Where the formula comes from, part one

Where the formula comes from, part two

Solve x² + 4x + 1 = 0 and test it

From the table to the curve

Draw your own parabola

First and second differences

Make a quadratic sequence

Where the line meets the curve

Test numbers in an inequality

Key words, A to Z