Chapter 6 · Higher tier

GCSE Maths Standard Form, Indices and Surds (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use index notation and the laws of indices to multiply, divide and raise powers to powers, with numbers and with letters
Work with zero and negative indices
Calculate with roots and (Higher) with fractional indices
Write very large and very small numbers in standard form and change them back
Compare, order and calculate with numbers in standard form, with and without a calculator
Give exact answers in terms of π
(Higher) Simplify, add, multiply and expand with surds, and rationalise denominators

You should already know

Multiplying and dividing whole numbers and decimals (Chapter 1)
Square numbers, cube numbers and roots (Chapter 2)
Fractions and reciprocals (Chapter 3)
Rounding to significant figures (Chapter 4)
Using letters in algebra

Where this leads

Expanding brackets and factorising (Chapter 7)
Area and circumference of circles (Chapter 18)
Pythagoras and exact trigonometry (Chapter 20)
Growth and decay (Chapters 5 and 25)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: N7, N8, N9, A4.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

6.1 Index notation and the laws of indices: Worksheet (PDF) · Answers (PDF) · lesson page
6.2 Zero and negative indices: Worksheet (PDF) · Answers (PDF) · lesson page
6.3 Roots and fractional indices: Worksheet (PDF) · Answers (PDF) · lesson page
6.4 Standard form: large and small numbers: Worksheet (PDF) · Answers (PDF) · lesson page
6.5 Comparing and calculating with standard form: Worksheet (PDF) · Answers (PDF) · lesson page
6.6 Exact answers with π: Worksheet (PDF) · Answers (PDF) · lesson page
6.7 Surds: simplifying and adding: Worksheet (PDF) · Answers (PDF) · lesson page
6.8 Multiplying, dividing and expanding with surds: Worksheet (PDF) · Answers (PDF) · lesson page
6.9 Rationalising the denominator and exact answers: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out each of these.

(a)2⁴

(b)3.7 × 1000

(d)√81 and ³√27

(e)Round 0.043756 to 3 significant figures.

Done

6.1

Index notation and the laws of indices

Edexcel 1MA1: N6, N7, A4
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Index notation and the laws of indicesComing soon

Fold a sheet of paper in half and it has 2 layers. Fold it again and there are 4 layers. Fold again: 8. Every fold doubles the number of layers, so after 3 folds there are 2 × 2 × 2 layers. Mathematicians write this in a short way: 2³, said “2 to the power 3”. This is called index notation.

In 3⁴ the base is 3 and the index is 4: four lots of 3 are multiplied together.

The big number is the base. The small raised number is the index (plural: indices; it is also called the power or exponent). The index tells you how many lots of the base are multiplied together. You already know two special cases: “squared” means index 2 and “cubed” means index 3. Be careful: 2³ is 2 × 2 × 2 = 8. It is not 2 × 3 = 6.

The index counts lots, and counting lots is easy. That is why there are three simple laws of indices. Each one comes straight from writing the powers out in full.

Law 1: 3⁴ × 3². Four lots of 3 and then two more lots of 3 makes 4 + 2 = 6 lots of 3.

Law 1 (multiplying). When you multiply powers of the same base, add the indices. 3⁴ × 3² = 3⁴⁺² = 3⁶.

Law 2 (dividing). When you divide powers of the same base, subtract the indices. 3⁵ ÷ 3² = (3 × 3 × 3 × 3 × 3) ÷ (3 × 3). Two of the 3s cancel on the top and bottom, leaving three of them: 3⁵⁻² = 3³.

Law 3 (power of a power). (3²)³ means three lots of 3², that is 3² × 3² × 3² = 3²⁺²⁺² = 3⁶. Three lots of 2 is 3 × 2, so we multiply the indices.

These laws only work when the bases are the same. They also work with letters, because a letter stands for a number. For example x² × x⁵ = x⁷. When there are numbers in front of the letters (the coefficients), deal with the numbers and the letters separately: 3x² × 4x⁵ = (3 × 4) × x²⁺⁵ = 12x⁷. A letter on its own has index 1, so x = x¹.

Key ideas

a^m × aⁿ = a^m+n (add the indices)
a^m ÷ aⁿ = a^m−n (subtract the indices)
(a^m)ⁿ = a^m×n (multiply the indices)
The bases must be the same for the first two laws.
With coefficients: multiply or divide the numbers, then use the laws on each letter. (2a)³ = 2³a³ = 8a³.
a¹ = a.

Worked example 6.1.1

Write 2 × 2 × 2 × 2 × 2 as a power of 2, and work out its value.

Method. Count how many 2s are multiplied to get the index, then multiply them out.

Count the 2s. The index of a power is the number of equal numbers being multiplied.
There are 5 lots of 2, so the index is 5.
Write it as a power. Base 2, index 5.
2 × 2 × 2 × 2 × 2 = 2⁵
Multiply step by step. Do one multiplication at a time.
2 × 2 = 4
4 × 2 = 8
8 × 2 = 16
16 × 2 = 32
Write the answer.
2⁵ = 32

Check. 2⁵ = 2² × 2³ = 4 × 8 = 32. ✓

Watch out. 2⁵ is not 2 × 5 = 10. The index counts lots of 2 multiplied together.

Worked example 6.1.2

Write 3⁴ × 3² as a single power of 3.

Method. When you multiply powers with the same base, you add the indices: a^m × aⁿ = a^m+n.

Check that the bases are the same. These laws only work when the base (the big number) is the same in both powers.
Both bases are 3.
Write each power out in full. This shows what is really happening: we are counting how many lots of the base are multiplied.
3⁴ = 3 × 3 × 3 × 3
3² = 3 × 3
Together: 4 lots of 3 and 2 lots of 3 make 6 lots of 3.
Add the indices. Joining the lots means adding the number of lots.
4 + 2 = 6
Write the answer as a single power. Keep the same base and use the new index.
3⁴ × 3² = 3⁶

Check. Work out the numbers: 3⁴ = 81 and 3² = 9. 81 × 9 = 729, and 3⁶ = 729. ✓

Watch out. Add the indices when you multiply. Do not multiply the indices, and do not multiply the bases (3² × 3⁴ is not 9⁶).

Worked example 6.1.3

Write (5⁷ × 5²) ÷ 5⁶ as a single power of 5.

Method. Use the laws of indices in order: add indices to multiply, subtract indices to divide.

Check that every base is the same. The laws of indices only work when the bases match.
Every base is 5.
Work out the top first: multiply, so add the indices. a^m × aⁿ = a^m+n.
7 + 2 = 9, so the top is 5⁹
Divide: subtract the indices. a^m ÷ aⁿ = a^m−n.
5⁹5⁶ = 5⁹ ÷ 5⁶ = 5³, because 9 − 6 = 3

Check. The total of the indices on top is 9 and on the bottom is 6; 9 − 6 = 3. ✓

Watch out. Deal with the multiplication in the top (and bottom) first, then divide. Only subtract once the top and bottom are each a single power.

Worked example 6.1.4

Work out (2³)⁴ as a power of 2, and then as a number.

Method. To raise a power to a power, multiply the indices: (a^m)ⁿ = a^m×n.

Write the bracket out as repeated multiplication. The outer index 4 means 4 lots of 2³ multiplied together.
(2³)⁴ = 2³ × 2³ × 2³ × 2³
= 2^{3 + 3 + 3 + 3}
That is 4 lots of 3, which is 4 × 3 = 12.
Multiply the indices. Each of the outer lots holds m of the base, so the total is m × n.
3 × 4 = 12
Write the answer as a single power. Keep the same base and use the new index.
(2³)⁴ = 2¹²
Work out the value. 2¹² means 12 lots of 2 multiplied together.
2¹² = 2 × 2 × … × 2 (12 lots of 2) = 4096

Check. Work out the numbers: 2³ = 8, and 8⁴ = 4096. Also 2¹² = 4096. ✓

Watch out. A power of a power: multiply the indices. Adding them would be for 3² × 3⁴, which is a different question.

Worked example 6.1.5

Simplify 3x² × 4x⁵.

Key words. In 3⁴, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 3⁴ = 3 × 3 × 3 × 3 = 81. A coefficient is the number in front of a letter, so in 7x³ the coefficient is 7. A letter on its own, like x, really means x¹.

Method. Multiply the numbers, then add the indices of each letter: a^m × aⁿ = a^m+n.

Multiply the numbers in front (the coefficients). The numbers can be dealt with separately from the letters.
3 × 4 = 12
Deal with the letter x: add the indices. The letter is the same in both terms, so the index law applies. (x on its own means x¹.)
x² × x⁵ = x^{2 + 5} = x⁷
Put the number and the letters together. Write the coefficient first, then the letters in alphabetical order.
12x⁷

Check. Choose numbers: x = 2. The original expression has the value 1536, and the answer 12x⁷ also has the value 1536. ✓

Watch out. Add the indices of the same letter only. You cannot combine different letters: a² × b³ stays as it is.

Worked example 6.1.6

Simplify (2a³)⁴.

Method. Raise the number and every letter to the power. Multiply the indices: (a^m)ⁿ = a^m×n.

Raise the number in front to the power. Everything inside the bracket is raised to the power, and that includes the number.
(2)⁴ = 2 × 2 × 2 × 2 = 16
Raise the letter a to the power: multiply the indices. (a^m)ⁿ = a^m×n.
(a³)⁴ = a^{3 × 4} = a¹²
Put it all together.
16a¹²

Check. Choose numbers: a = 2. The original expression has the value 65,536, and the answer 16a¹² also has the value 65,536. ✓

Watch out. The number in front is raised to the power too: (2a)³ = 8a³, not 2a³.

Worked example 6.1.7

Simplify (2x³y)³ × 4xy² ÷ 8x²y.

Method. Work through the expression one operation at a time, using the laws of indices on the number and on each letter.

Look at the order of the operations. Work from the left, one operation at a time, exactly as in BIDMAS (brackets and powers first).
Do the first operation: raise to the power 3. Raise the number and every index to the power: multiply the indices.
(2)³ = 2 × 2 × 2 = 8
(x³)³ = x^{3 × 3} = x⁹
(y¹)³ = y^{1 × 3} = y³
Result so far: 8x⁹y³
Do the second operation: multiply by 4xy². Multiply the numbers; add the indices of each letter.
8 × 4 = 32
x⁹ × x¹ = x^{9 + 1} = x¹⁰
y³ × y² = y^{3 + 2} = y⁵
Result so far: 32x¹⁰y⁵
Do the third operation: divide by 8x²y. Divide the numbers; subtract the indices of each letter.
32 ÷ 8 = 4
x¹⁰ ÷ x² = x^{10 − 2} = x⁸
y⁵ ÷ y¹ = y^{5 − 1} = y⁴
Result so far: 4x⁸y⁴

Check. Choose numbers: x = 2, y = 3. The original expression gives 82,944 and the answer 4x⁸y⁴ gives 82,944. ✓

Watch out. Keep the working in the same order each time. Deal with brackets and powers first, then multiplication and division from left to right.

Watch out

Do not multiply the bases and do not multiply the indices when you are multiplying powers. 3² × 3⁴ is not 9⁶. The base stays 3 and the indices add: 3⁶. And you cannot combine powers of different letters or different numbers: x³ + x² does not equal x⁵, and 2³ × 3² is not 6⁵.

Exam tip

In an exam, show the indices adding: write 3⁴⁺² = 3⁶. If the question says “work out the value” you must also give the final number.

6.2

Zero and negative indices

Edexcel 1MA1: N7, A4
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Zero and negative indicesComing soon

So far every index has been a positive whole number. But what do 2⁰ and 2⁻¹ mean? We cannot multiply “zero lots of 2”, so instead we follow the pattern.

Going one step to the right the index goes down by 1 and the value is divided by 2. Keep the pattern going and 2⁰ = 1, 2⁻¹ = 1/2, 2⁻² = 1/4.

Look at the boxes from left to right: 8, 4, 2, then the next number must be 2 ÷ 2 = 1. So 2⁰ = 1. The same thing happens with any base, so any non-zero number to the power 0 equals 1. You can also see it from the division law: a³ ÷ a³ = a³⁻³ = a⁰, but anything divided by itself is 1.

Continue the pattern past 1: the next value is 1 ÷ 2 = 1/2, so 2⁻¹ = 1/2. Then 2⁻² = 1/4 and 2⁻³ = 1/8. A negative index means “one over” (the reciprocal; see Chapter 3):

a⁻ⁿ = 1aⁿ

A negative index does not make the answer negative. It makes the answer small: a fraction. For a fraction base, the reciprocal turns the fraction upside down, so 23⁻² = 32² = 94.

Two more things to know. 10⁻¹ = 0.1, 10⁻² = 0.01 and 10⁻³ = 0.001 (you will need these for standard form in section 6.4). And all the laws of section 6.1 still work with zero and negative indices. For example 5² × 5⁻⁴ = 5²⁺⁽⁻⁴⁾ = 5⁻² = 125, and x³ ÷ x⁵ = x⁻² = 1x².

Key ideas

a⁰ = 1 for any number a that is not 0. (5⁰ = 1, 1000⁰ = 1.)
a¹ = a.
a⁻ⁿ = 1 ÷ aⁿ. A negative index means “one over”, not a negative number.
(a/b)⁻ⁿ = (b/a)ⁿ: flip the fraction and make the index positive.
10⁻¹ = 0.1, 10⁻² = 0.01, 10⁻³ = 0.001.
The laws of indices work for all whole-number indices, including zero and negative ones.

Worked example 6.2.1

Work out 5⁰ + 2 × 3⁰.

Method. A number (or bracket) to the power zero is 1.

Remember the rule for a zero index. Any non-zero number raised to the power 0 equals 1. This is because aⁿ ÷ aⁿ = 1, and by the division law it is also aⁿ⁻ⁿ = a⁰.
a⁰ = 1
Replace every power with index 0 by 1, and work out the rest. Powers are worked out before multiplying and adding (BIDMAS): each zero power becomes 1.
5⁰ = 1 and 3⁰ = 1
5⁰ + 2 × 3⁰ = 1 + 2 × 1
= 1 + 2 = 3
Write the answer.
3

Check. Look again: each 0 index was changed to 1 and nothing else was changed. ✓

Watch out. 5⁰ = 1, not 0 and not 5. And in 5x⁰ only the x has the index 0, so it is 5 × 1 = 5.

Worked example 6.2.2

Work out 2⁻³.

Method. A negative index means "one over": a⁻ⁿ = 1 ÷ aⁿ. For a fraction, flip it upside down and make the index positive.

Spot the negative index. A negative index does not make the answer negative. It means "one over".
2⁻³: the index is −3
Turn it into a reciprocal with a positive index. The rule is a⁻ⁿ = 1aⁿ. (Check with a pattern: 2² = 4, 2¹ = 2, 2⁰ = 1, and each step is ÷ 2, so 2⁻¹ = 12.)
2⁻³ = 12³
Work out the bottom. 2³ means 3 lots of 2 multiplied together.
2³ = 2 × 2 × 2 = 8
Write the answer as a fraction.
2⁻³ = 18

Check. Multiply the answer by 2³: 18 × 8 = 1, and 2⁻³ × 2³ = 2⁰ = 1. ✓

Watch out. A negative index does not give a negative answer: 2⁻³ = 18, not a negative number.

Worked example 6.2.3

Work out (23)⁻².

Method. A negative index means "one over": a⁻ⁿ = 1 ÷ aⁿ. For a fraction, flip it upside down and make the index positive.

Spot the negative index. A negative index means "reciprocal". The reciprocal of a fraction is the fraction turned upside down.
(23)⁻²: the index is −2
Flip the fraction and change the index to positive. Negative index = reciprocal. The reciprocal of 23 is 32.
(23)⁻² = (32)²
Raise the top and the bottom to the power. The index 2 applies to the numerator and to the denominator.
3²2² = 94
Write the answer (simplify if possible).
94

Check. Multiply the answer by (23)² = 49: the product is 1, as (23)⁻² × (23)² = (23)⁰ = 1. ✓

Watch out. A negative index does not give a negative answer: (23)⁻² = 94, not a negative number.

Worked example 6.2.4

Write 125 as a power of 5.

Method. Write the bottom of the fraction as a power, then use a⁻ⁿ = 1/aⁿ backwards.

Look at the fraction. It has 1 on the top and a number on the bottom, so it is "one over" something. A negative index means "one over".
125
Write the denominator as a power of the base. 25 = 5 × 5 = 5²
125 = 15²
Use the rule for a negative index. "One over a power" is the same as the power with a negative index: 1aⁿ = a⁻ⁿ.
15² = 5⁻²

Check. Convert back: 5⁻² = 15² = 125. ✓

Watch out. The number on the bottom becomes the base. The index gets the minus sign.

Worked example 6.2.5

Work out 5² × 5⁻⁴, giving your answer as a fraction.

Method. When you multiply powers with the same base, you add the indices: a^m × aⁿ = a^m+n.

Check that the bases are the same. These laws only work when the base (the big number) is the same in both powers.
Both bases are 5.
Say what the rule means. The index counts the lots of the base. Multiplying powers joins the lots together.
5² has 2 lots of 5; 5⁻⁴ has −4 lots of 5.
Add the indices. Joining the lots means adding the number of lots. The rule works for negative indices too, so take care with the signs.
2 + (−4) = −2
Write the answer as a single power. Keep the same base and use the new index.
5² × 5⁻⁴ = 5⁻²
Work out the value. A negative index means "one over" the positive power (see section 6.2).
5⁻² = 15² = 125

Watch out. Add the indices when you multiply. Do not multiply the indices, and do not multiply the bases (3² × 3⁴ is not 9⁶).

Worked example 6.2.6

Simplify x³ ÷ x⁵, and write your answer with a positive index.

Method. Divide the numbers, then subtract the indices of each letter: a^m ÷ aⁿ = a^m−n.

Deal with the letter x: subtract the indices. The letter is the same in both terms, so the index law applies. (x on its own means x¹.)
x³ ÷ x⁵ = x^{3 − 5} = x⁻² = 1x²
Put the number and the letters together. Write the coefficient first, then the letters in alphabetical order.
x⁻²
Write with positive indices. A negative index means "one over": a⁻ⁿ = 1/aⁿ.
x⁻² = 1x²

Check. Choose numbers: x = 2. The original expression has the value 14, and the answer x⁻² also has the value 14. ✓

Watch out. Subtract the indices (top minus bottom), and divide the numbers. Do not subtract the numbers.

Worked example 6.2.7

Simplify (2x⁻³)², writing your answer with a positive index.

Method. Raise the number and every letter to the power. Multiply the indices: (a^m)ⁿ = a^m×n.

Raise the number in front to the power. Everything inside the bracket is raised to the power, and that includes the number.
(2)² = 2 × 2 = 4
Raise the letter x to the power: multiply the indices. (a^m)ⁿ = a^m×n.
(x⁻³)² = x^{−3 × 2} = x⁻⁶
Put it all together.
4x⁻⁶
Write with positive indices. A negative index means "one over": a⁻ⁿ = 1/aⁿ.
4x⁻⁶ = 4x⁶

Check. Choose numbers: x = 2. The original expression has the value 116, and the answer 4x⁻⁶ also has the value 116. ✓

Watch out. The number in front is raised to the power too: (2a)³ = 8a³, not 2a³.

Watch out

A negative index does not make a number negative. 2⁻³ is 18, not −8 and not −6. Also be careful with zero: 5⁰ = 1, not 0. And in 5x⁰ the index 0 belongs to the x only, so it is 5 × 1 = 5, while (5x)⁰ = 1.

Exercise 6D Index equations and mixed practice

Non-calculator

FFind the value of n in each equation.

(a)2ⁿ = 32

(b)3ⁿ = 81

(d)10ⁿ = 1000

(e)2ⁿ = 128

Done

FFind the value of n in each equation. (The answers are negative whole numbers.)

(a)2ⁿ = 18

(b)3ⁿ = 19

(d)5ⁿ = 15

(e)4ⁿ = 116

Done

FFind the value of x.

(a)2^{x + 3} = 2⁸

(b)3^{x − 1} = 3⁴

(d)10^{x − 2} = 10³

Done

FWork out each. Give your answer as a fraction.

(a)2⁻¹ + 2⁻²

(b)3⁻¹ + 3⁻²

Done

PSWrite these numbers in order, smallest first: 2⁻¹, 2⁰, 2⁻³, 2².

Done

RWrite 18 as a power of 2, and write 8 × 132 as a power of 2.

(a)18

(b)8 × 132

Done

RShow that (13)⁻² × 3⁻⁴ = 19.

Done

RFind the value of x when 2^x × 2³ ÷ 2⁻² = 2⁹.

Done

6.3

Roots and fractional indices

Edexcel 1MA1: N6, N7, A4
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Roots and fractional indicesComing soon

A root undoes a power. A square has area 25 cm². Its side is the number that multiplies by itself to give 25. That number is 5, so the square root of 25 is 5. We write √25 = 5. A cube has volume 27 cm³. Its edge is the number that gives 27 when multiplied by itself three times, so the cube root of 27 is 3, written ³√27 = 3.

Roots find the side of a square from its area and the edge of a cube from its volume.

Learn the perfect squares up to 15² = 225 and the cubes 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000 (see Chapter 2). Then roots of these are quick. You can also take roots of fractions and decimals: √(9/16) = 3/4 because (3/4)² = 9/16, and √0.25 = 0.5 because 0.5 × 0.5 = 0.25.

Most numbers are not perfect squares, but we can say where their roots lie. √50 is between √49 = 7 and √64 = 8, so it is between 7 and 8. A calculator gives √50 = 7.07…. Use the √ key for square roots and the cube root key (often SHIFT then √) for cube roots.

Remember: (√a)² = a. Equations such as x² = 49 have two solutions, x = 7 and x = −7, because (−7)² = 49 too. The √ sign itself means the positive root. You cannot take the square root of a negative number (no real number squares to −9), but a cube root of a negative number is fine: ³√(−8) = −2.

Key ideas

√a is the positive number that multiplies by itself to give a. ³√a multiplies three times.
(√a)² = a, and x² = k has two solutions, ±√k.
To place a root between two whole numbers use the perfect squares (or cubes) on each side.
Roots of fractions: take the root of the top and of the bottom.

Worked example 6.3.1

Work out √144.

Key words. The square root of a number undoes squaring: √49 = 7 because 7² = 49. The cube root undoes cubing: ³√27 = 3 because 3³ = 27.

Method. Find the number that gives the original when raised to the power 2.

Say what the question is asking. The square root of a number is the number that gives the original number when 2 lots of it are multiplied together. It undoes the power 2.
√144: which number to the power 2 equals 144?
Test the powers in order. List the powers until you reach the number.
1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100, 11² = 121, 12² = 144
Write the answer. 12² = 144.
12

Check. Check: 12² = 12 × 12 = 144. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.2

Work out √916 and √0.25.

Key words. The square root of a number undoes squaring: √49 = 7 because 7² = 49. The cube root undoes cubing: ³√27 = 3 because 3³ = 27.

Method. Find the number that gives the original when raised to the power 2.

Spot that the number is a fraction. A fraction is a top number over a bottom number, and we can deal with each separately.
916: top 9, bottom 16
Take the root of the top and of the bottom. For a fraction, √(a ÷ b) = √a ÷ √b.
√9 = 3 and √16 = 4
Write the answer as a fraction.
34

Check. Check: 34 to the power 2 = 916. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.3

Write down the cube root of 125.

Key words. The square root of a number undoes squaring: √49 = 7 because 7² = 49. The cube root undoes cubing: ³√27 = 3 because 3³ = 27.

Method. Find the number that gives the original when raised to the power 3.

Say what the question is asking. The cube root of a number is the number that gives the original number when 3 lots of it are multiplied together. It undoes the power 3.
³√125: which number to the power 3 equals 125?
Test the powers in order. List the powers until you reach the number.
1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125
Write the answer. 5³ = 125.
5

Check. Check: 5³ = 5 × 5 × 5 = 125. ✓

Watch out. The square root of 49 is 7, not 24.5. Do not halve the number: a root undoes a power.

Worked example 6.3.4

Between which two whole numbers does √50 lie?

Key words. A perfect square is the square of a whole number (1, 4, 9, 16…). A perfect cube is the cube of a whole number (1, 8, 27, 64…).

Method. Find the two perfect squares that the number lies between, then take roots.

Find the perfect powers on either side of the number. We know the square root of perfect squares exactly, so use them as markers.
7² = 49 and 8² = 64
Place the number between them. 50 lies between 49 and 64.
49 < 50 < 64
Take the root of each part. Roots keep the order (a bigger number has a bigger root).
7 < √50 < 8

Check. A calculator gives √50 = 7.071, which is between 7 and 8. ✓

Watch out. Between two whole numbers means one is below and the next one is above.

Worked example 6.3.5

Work out √36 + ³√27 − 2³.

Method. Work out each root or power separately, then add and subtract.

Work out each root or power on its own. A root undoes a power; a power is repeated multiplication.
√36 = 6, because 6 × 6 = 36
³√27 = 3, because 3 × 3 × 3 = 27
2³ = 2 × 2 × 2 = 8
Put the values into the calculation. Keep the signs the same.
√36 + ³√27 − 2³ = 6 + 3 − 8
Work out the answer. Add and subtract from left to right.
6 + 3 − 8 = 1

Check. Work through it again from the left: 6 + 3 − 8 = 1. ✓

Watch out. Do each root or power first (BIDMAS: powers and roots before adding and subtracting).

Watch out

The square root of 36 is 6, not 18. A root does not halve the number. And √(9 + 16) is not √9 + √16: √25 = 5 but 3 + 4 = 7. Add or subtract inside the root first.

Higher tier: fractional indices

The laws of indices tell us what a fractional index must mean. Take 25^1/2 and multiply it by itself: 25^1/2 × 25^1/2 = 25^1/2+1/2 = 25¹ = 25. So 25^1/2 is a number that multiplies by itself to give 25. That is the square root: 25^1/2 = √25 = 5. In the same way a^1/3 is the cube root of a, and a^1/n is the nth root of a.

For a general fraction, the denominator is the root and the numerator is the power: a^m/n = (ⁿ√a)^m. For example 8^2/3 = (³√8)² = 2² = 4. Take the root first: it keeps the numbers small. A negative fractional index combines both ideas: take the reciprocal, then the root and power. 16^−3/4 = (1/16)^3/4 = (1/2)³ = 1/8.

The same laws work with letters: (16x⁸)^3/4 = 16^3/4 × x^8×3/4 = 8x⁶. And you can solve equations: to undo a power p/q raise both sides to the power q/p. If the bases differ, write both as powers of the same number: 4^x = 8 becomes (2²)^x = 2³, so 2x = 3 and x = 3/2.

Key ideas (Higher)

a^1/2 = √a, a^1/3 = ³√a, a^1/n = nth root of a.
a^m/n = (ⁿ√a)^m: denominator is the root, numerator is the power. Do the root first.
a^−m/n = 1 ÷ a^m/n: flip first.
To solve x^p/q = k, raise both sides to the power q/p.

Worked example 6.3.6

Work out 8^2/3.

Key words. In 3⁴, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 3⁴ = 3 × 3 × 3 × 3 = 81. A root undoes a power: the square root of 49 is 7 because 7² = 49.

Method. Fractional index: the denominator is the root, the numerator is the power. a^m/n = (ⁿ√a)^m. Do the root first.

Read the fractional index. The denominator of the index tells you which root to take. The numerator tells you which power to raise to. For example 8^1/3 is the cube root of 8, because (8^1/3)³ = 8¹ = 8.
8^2/3: denominator 3 means the cube root; numerator 2 means the power 2
Take the cube root first. Doing the root first keeps the numbers small. Ask: which number multiplied by itself 3 times gives the base?
cube root of 8: 2³ = 8, so it is 2.
Raise the answer to the power 2. The numerator is the power.
2² = 2 × 2 = 4

Check. Check with the other order: raise to the power first, then take the root. 8² = 64 and its cube root is 4. ✓

Watch out. a^1/2 is the square root of a. It is not a divided by 2 or a × one half.

Worked example 6.3.7

Work out 16^−3/4.

Key words. In 3⁴, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 3⁴ = 3 × 3 × 3 × 3 = 81. A root undoes a power: the square root of 49 is 7 because 7² = 49.

Method. Fractional index: the denominator is the root, the numerator is the power. a^m/n = (ⁿ√a)^m. Do the root first.

Read the fractional index. The denominator of the index tells you which root to take. The numerator tells you which power to raise to. For example 8^1/3 is the cube root of 8, because (8^1/3)³ = 8¹ = 8.
16^−3/4: denominator 4 means the fourth root; numerator 3 means the power 3; the minus sign means the reciprocal
Deal with the minus sign: take the reciprocal first. A negative index means "one over" (or flip the fraction). Doing this first lets us finish with a positive index.
16^−3/4 = (116)^3/4
Take the fourth root first. Doing the root first keeps the numbers small. Ask: which number multiplied by itself 4 times gives the base?
fourth root of 116: the fourth root of the top and of the bottom: 1 and 2, so it is 12.
Raise the answer to the power 3. The numerator is the power.
(12)³ = 18 = 18

Check. Check with the other order: raise to the power first, then take the root. 116³ = 14096 and its fourth root is 18. ✓

Watch out. a^1/2 is the square root of a. It is not a divided by 2 or a × one half.

Worked example 6.3.8

Work out (278)^−2/3.

Key words. In 3⁴, the base is 3 and the index (also called the power or exponent) is 4. The index tells you how many lots of the base are multiplied together: 3⁴ = 3 × 3 × 3 × 3 = 81. A root undoes a power: the square root of 49 is 7 because 7² = 49.

Method. Fractional index: the denominator is the root, the numerator is the power. a^m/n = (ⁿ√a)^m. Do the root first.

Read the fractional index. The denominator of the index tells you which root to take. The numerator tells you which power to raise to. For example 8^1/3 is the cube root of 8, because (8^1/3)³ = 8¹ = 8.
(278)^−2/3: denominator 3 means the cube root; numerator 2 means the power 2; the minus sign means the reciprocal
Deal with the minus sign: take the reciprocal first. A negative index means "one over" (or flip the fraction). Doing this first lets us finish with a positive index.
(278)^−2/3 = (827)^2/3
Take the cube root first. Doing the root first keeps the numbers small. Ask: which number multiplied by itself 3 times gives the base?
cube root of 827: the cube root of the top and of the bottom: 2 and 3, so it is 23.
Raise the answer to the power 2. The numerator is the power.
(23)² = 49 = 49

Check. Check with the other order: raise to the power first, then take the root. 827² = 64729 and its cube root is 49. ✓

Watch out. a^1/2 is the square root of a. It is not a divided by 2 or a × one half.

Worked example 6.3.9

Simplify (16x⁸)^3/4.

Method. Raise the number and every letter to the power. Multiply the indices: (a^m)ⁿ = a^m×n.

Raise the number in front to the power. Everything inside the bracket is raised to the power, and that includes the number.
The 4th root of 16 is 2, because 2⁴ = 16.
Then 2³ = 8
Raise the letter x to the power: multiply the indices. (a^m)ⁿ = a^m×n.
(x⁸)^3/4 = x^{8 × 3/4} = x⁶
Put it all together.
8x⁶

Check. Choose numbers: x = 4. The original expression has the value 32,768, and the answer 8x⁶ also has the value 32,768. ✓

Watch out. The number in front is raised to the power too: (2a)³ = 8a³, not 2a³.

Worked example 6.3.10

Solve x^3/2 = 27.

Method. Undo a fractional power by raising both sides to the reciprocal power.

Spot the fractional index. The letter is raised to the power p/q. To get the letter on its own we need to undo that power.
x^3/2 = 27
Raise both sides to the reciprocal power. The reciprocal of 3/2 is 2/3. Raising to this power gives (x^3/2)^2/3 = x¹, which is just x. Do the same to the right-hand side to keep the equation balanced.
x = 27^2/3
Evaluate: root first, then power. The denominator 3 is the root and the numerator 2 is the power.
the cube root of 27 is 3, then 3² = 9
x = 9

Check. Put x = 9 back: the square root of 9 is 3, and 3³ = 27. ✓

Watch out. Use the reciprocal of the index (flip the fraction), not the negative of it.

Worked example 6.3.11

Solve 4^x = 8.

Method. Write both sides as powers of the same base, then compare the indices.

Write both sides as powers of the same base. The bases 4 and 8 look different, but both are powers of 2.
4 = 2²
8 = 2³
Replace them in the equation. Use the power-of-a-power law: (a^m)ⁿ = a^m×n. So raising 2² to the power x multiplies the indices.
(2²)^x = 2^2x, so 2^2x = 2³
Equate the indices. The bases are the same, so the indices must be equal.
2x = 3
Solve. Divide both sides by 2.
x = 32 = 32

Check. Put x = 32 back: 4^3/2 = (2²)^3/2 = 2^{2 × (3/2)} = 2³ = 8. ✓

Watch out. Find the common base first (2, 3 or 5 are the usual ones). Then the powers-of-a-power law turns the problem into a simple equation.

Watch out

a^1/2 is the square root of a. It is not a ÷ 2. And a^2/3 needs both the cube root and the square: forgetting the numerator is a very common error.

Exercise 6F Powers and roots with a calculator

Calculator allowed

FUse a calculator to work out each root. Give your answers to 3 significant figures.

(a)√7

(b)√130

(c)√2000

(d)³√50

(e)³√300

Done

FUse a calculator to work out each power. Give your answers to 3 significant figures.

(a)2.5⁴

(b)1.05¹⁰

(d)0.9⁵

(e)8⁻²

Done

PSA square picture has an area of 90 cm². Work out the length of one side to 1 decimal place.

Done

PS£1500 is invested. After 5 years it is worth £1500 × 1.04⁵. Work out its value, to the nearest penny.

Done

PSWork out √(5.2² + 3.9²).

Done

FUse a calculator to work out each value. Give your answers to 3 significant figures.

(a)√11

(b)√59

(d)0.5⁻³

Done

RShow, using a calculator, that ³√100 is between 4.6 and 4.7.

Done

6.4

Standard form: large and small numbers

Edexcel 1MA1: N9, N7
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Standard form: large and small numbersComing soon

The Sun is about 150,000,000,000 m from the Earth. A hydrogen atom is about 0.0000000001 m wide. These numbers are long and easy to get wrong. Standard form is a short, standard way to write very big and very small numbers. It uses powers of 10 (see section 6.2).

Real things on a line of powers of 10. Each step along the line is a factor of 100.

A number is in standard form when it is written as A × 10ⁿ where

A is at least 1 and less than 10 (one non-zero digit before the decimal point), and
n is a whole number (positive, negative or zero).

Big numbers have a positive power; small numbers (less than 1) have a negative power. For example 150,000,000,000 = 1.5 × 10¹¹ and 0.0000000001 = 1 × 10⁻¹⁰.

To write a number in standard form: (1) put the decimal point after the first non-zero digit to make A; (2) count how many places the decimal point moved; (3) that count is the power. Moving to the left gives a positive power, and moving to the right gives a negative power. For example 45,000,000: the point moves 7 places left, so 4.5 × 10⁷. And 0.00034: the point moves 4 places right, so 3.4 × 10⁻⁴.

Going back: 3.2 × 10⁵ means move the decimal point 5 places right, filling with zeros, to get 320,000.

To change back to an ordinary number do the opposite: a positive power moves the decimal point right, a negative power moves it left, one place for each power of 10. The number is not made negative by a negative power: 4 × 10⁻² = 0.04.

Sometimes a number looks like standard form but the front number is not between 1 and 10, for example 45 × 10³. Fix it by changing A and the power together: 45 = 4.5 × 10, so 45 × 10³ = 4.5 × 10 × 10³ = 4.5 × 10⁴. On a calculator, standard form is often shown with an E: 3.6E8 means 3.6 × 10⁸.

Key ideas

Standard form: A × 10ⁿ with 1 ≤ A < 10 and n a whole number.
Big number: positive power. Number between 0 and 1: negative power.
Count how many places the decimal point moves. Left means positive, right means negative.
To fix 45 × 10³: write 45 as 4.5 × 10, then add 1 to the power.
A calculator shows 3.6 × 10⁸ as 3.6E8.

Worked example 6.4.1

Write 45,000,000 in standard form.

Key words. Standard form is a way to write very big or very small numbers as A × 10ⁿ, where A is at least 1 and less than 10 (so exactly one non-zero digit before the decimal point) and n is a whole number (positive, negative or zero).

Method. Write the number as A × 10ⁿ: put the decimal point after the first non-zero digit, then count how many places it moved.

Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
The first non-zero digit of 45,000,000 is 4.
Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
A = 4.5
Count how many places the decimal point has moved. Starting from 45000000 the decimal point has to move 7 places to the left to sit after the 4. Each place moved is a factor of 10.
45000000 → 4.5: 7 places to the left
Decide the sign of the power. The number is 10 or more (a big number). We made it smaller to get A, so we must multiply by 10 to put the size back.
n is positive: n = 7
Write the answer. Put A and the power of 10 together.
4.5 × 10⁷

Check. Convert back: 4.5 × 10⁷ = 45000000 = 45,000,000. ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Worked example 6.4.2

Write 0.00034 in standard form.

Method. Write the number as A × 10ⁿ: put the decimal point after the first non-zero digit, then count how many places it moved.

Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
The first non-zero digit of 0.00034 is 3.
Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
A = 3.4
Count how many places the decimal point has moved. Starting from 0.00034 the decimal point has to move 4 places to the right to sit after the 3. Each place moved is a factor of 10.
0.00034 → 3.4: 4 places to the right
Decide the sign of the power. The number is less than 1 (a small number). We made it bigger to get A, so we must divide by 10 to put the size back. Dividing by 10 is the same as multiplying by 10⁻¹.
n is negative: n = −4
Write the answer. Put A and the power of 10 together.
3.4 × 10⁻⁴

Check. Convert back: 3.4 × 10⁻⁴ = 0.00034 = 0.00034. ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Worked example 6.4.3

Write 3.07 × 10⁴ as an ordinary number.

Method. A positive power of 10 makes the number bigger (move the decimal point right). A negative power makes it smaller (move the decimal point left).

Look at the power of 10. The power is 4, which is positive, so the number is big. Multiplying by 10 moves every digit one place to the left, which is the same as moving the decimal point one place to the right.
10⁴ means multiply by 10, 4 times
Move the decimal point 4 places to the right. Move it one place at a time. If there are not enough digits, fill the gaps with zeros.
3.07 → 30.7 → 307 → 3070 → 30700
Write the answer. The last number in the chain is the answer.
30,700

Check. Convert back to standard form: 30700 has first non-zero digit 3, so A = 3.07 and n = 4. ✓

Watch out. A negative power makes a number smaller, but it does not make it negative.

Worked example 6.4.4

Write 4.1 × 10⁻⁵ as an ordinary number.

Method. A positive power of 10 makes the number bigger (move the decimal point right). A negative power makes it smaller (move the decimal point left).

Look at the power of 10. The power is −5, which is negative, so the number is small. Dividing by 10 moves every digit one place to the right, which is the same as moving the decimal point one place to the left.
10⁻⁵ means divide by 10, 5 times
Move the decimal point 5 places to the left. Move it one place at a time and fill the new places with zeros.
4.1 → 0.41 → 0.041 → 0.0041 → 0.00041 → 0.000041
Write the answer. The last number in the chain is the answer.
0.000041

Check. Convert back to standard form: 0.000041 has first non-zero digit 4, so A = 4.1 and n = −5. ✓

Watch out. A negative power makes a number smaller, but it does not make it negative.

Worked example 6.4.5

Write 45 × 10³ in standard form.

Method. Change the front number A so that it is between 1 and 10, and adjust the power of 10 to match.

Check the rule for standard form. In A × 10ⁿ, A must be at least 1 and less than 10.
A = 45, which is 10 or more, so this is not in standard form.
Write 45 as a number between 1 and 10 times a power of 10. Moving the decimal point 1 place to the left makes the number 10 times smaller, so we must multiply by 10¹ to keep the value the same.
45 = 4.5 × 10¹
Replace A in the original and combine the powers of 10. Powers of 10 multiply by adding the indices: 10^a × 10^b = 10^a+b.
45 × 10³ = 4.5 × 10¹ × 10³
= 4.5 × 10^{1 + 3}
Write the answer.
4.5 × 10⁴

Check. Both forms are the same number, 45000: 45 × 1000 = 45000 and 4.5 × 10000 = 45000. ✓

Watch out. If you make A smaller by moving the point left, the power of 10 gets bigger by the same number of places (and the other way round).

Worked example 6.4.6

Write 299,792,458 in standard form correct to 3 significant figures.

Method. Write the number as A × 10ⁿ: put the decimal point after the first non-zero digit, then count how many places it moved.

Round to 3 significant figures first. Significant figures count from the first non-zero digit.
299,792,458 = 300,000,000 to 3 s.f.
Find the first non-zero digit. In standard form A is at least 1 and less than 10, so A has exactly one non-zero digit in front of the decimal point. That digit is the first non-zero digit of the number.
The first non-zero digit of 300,000,000 is 3.
Place the decimal point just after it to make A. The digits of A are the same as the digits of the number (without the zeros at the ends).
A = 3.00
Count how many places the decimal point has moved. Starting from 300000000 the decimal point has to move 8 places to the left to sit after the 3. Each place moved is a factor of 10.
300000000 → 3.00: 8 places to the left
Decide the sign of the power. The number is 10 or more (a big number). We made it smaller to get A, so we must multiply by 10 to put the size back.
n is positive: n = 8
Write the answer. Put A and the power of 10 together.
3.00 × 10⁸

Check. Convert back: 3.00 × 10⁸ = 300000000 (which is 300,000,000 to 3 s.f.). ✓

Watch out. A big number has a positive power and a small number (less than 1) has a negative power. Do not count the zeros, count the places the point moves.

Watch out

Count how far the decimal point moves, not how many zeros you can see. And check the sign: a number less than 1 must have a negative power. 0.0072 = 7.2 × 10⁻³, not 7.2 × 10³.

Exam tip

A is always at least 1 and less than 10. If you get 12 × 10⁵ or 0.7 × 10⁴ you are not finished: it is not standard form yet.

Exercise 6J Standard form in context

Non-calculator

PSWrite each measurement in standard form.

(a)The diameter of the Sun is 1,392,000 km.

(b)The wavelength of red light is 0.0000007 m.

(c)The population of the UK is about 67,000,000.

(d)A human hair is about 0.00007 m wide.

(e)The nearest star is about 40,000,000,000,000 km away.

Done

PSWrite each of these as an ordinary number.

(a)The Earth is 1.5 × 10⁸ km from the Sun.

(b)A virus is 1.2 × 10⁻⁷ m long.

(c)A computer can do 3 × 10⁹ operations per second.

(d)A grain of pollen has a mass of 2.5 × 10⁻⁹ kg.

Done

FWrite each of these numbers in standard form.

(a)85,000

(b)0.0092

(d)0.00000058

(e)1,200,000,000

Done

FWrite each number as an ordinary number.

(a)6.3 × 10⁴

(b)9 × 10⁻³

(d)4.4 × 10⁻²

Done

FWrite each number in standard form.

(a)32 × 10⁶

(b)0.45 × 10³

(d)0.09 × 10⁴

Done

RZoe says: “50 × 10⁴ is bigger than 5 × 10⁴, so it must be in standard form.” Is she right? Explain.

Done

PSA scientist measures a bacterium as 0.0000025 m long. She writes 25 × 10⁻⁷ m. Write this correctly in standard form.

Done

RShow that 3 × 10⁵ is 100 times bigger than 3 × 10³.

Done

6.5

Comparing and calculating with standard form

Edexcel 1MA1: N9
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Comparing and calculating with standard formComing soon

Standard form makes big numbers easy to compare. Look at the powers of 10 first. A bigger power means a bigger number, because the front numbers are all between 1 and 10 and cannot make up the difference. So 3.2 × 10⁵ is bigger than 9.9 × 10⁴. If the powers are the same, compare the front numbers. Be careful with negative powers: −3 is bigger than −5, so 10⁻³ = 0.001 is bigger than 10⁻⁵ = 0.00001.

Standard form makes it easy to calculate with huge numbers, such as the time light takes to reach us from the Sun.

Multiplying. Multiply the front numbers, add the powers of 10 (law 1 of indices), then check that A is between 1 and 10. (2 × 10³) × (3 × 10⁴) = 6 × 10⁷. If the front numbers multiply to 10 or more, adjust: (4 × 10⁵) × (3 × 10²) = 12 × 10⁷ = 1.2 × 10⁸.

Dividing. Divide the front numbers and subtract the powers. If A comes out less than 1, adjust: (3 × 10⁵) ÷ (5 × 10⁻²) = 0.6 × 10⁷ = 6 × 10⁶.

Adding and subtracting. The powers of 10 must be the same first (like fractions needing a common denominator). 3.2 × 10⁴ + 8.5 × 10³ = 3.2 × 10⁴ + 0.85 × 10⁴ = 4.05 × 10⁴.

With a calculator use the ×10^x key (sometimes EXP), then write the answer in standard form to the accuracy asked for. Always estimate first by rounding each A to a whole number. In real problems decide which operation is needed, for example time = distance ÷ speed.

Higher tier: you may also need to find a power or a square root of a number in standard form: (2 × 10³)³ = 2³ × 10⁹ = 8 × 10⁹. Raise A to the power and multiply the index of 10 by the power.

Key ideas

Compare the powers of 10 first, then the front numbers.
Multiply: multiply the A values and add the powers. Divide: divide the A values and subtract the powers.
Add or subtract: make the powers the same first.
After any calculation check that A is at least 1 and less than 10; if not, adjust A and the power together.
Calculator: use the ×10^x key and check with an estimate.

Worked example 6.5.1

Which is larger, 3.2 × 10⁵ or 9.9 × 10⁴?

Method. Compare the powers of 10 first. If the powers are the same, compare the front numbers.

Check both numbers are in standard form. The front numbers must be between 1 and 10 for this comparison to work.
3.2 and 9.9 are both at least 1 and less than 10.
Compare the powers of 10 first. A bigger power of 10 means a bigger number, because both front numbers are between 1 and 10 and so cannot make up the difference.
10⁵ and 10⁴: 5 > 4
Decide. The powers are different, so the powers decide it. We do not need to look at the front numbers.
3.2 × 10⁵ > 9.9 × 10⁴

Check. As ordinary numbers: 320000 and 99000, and 320000 is bigger than 99000. ✓

Watch out. Do not compare the front numbers first. 9.9 × 10² is much smaller than 1.1 × 10⁵.

Worked example 6.5.2

Write these in order, smallest first: 2.5 × 10⁴, 7.1 × 10⁻³, 2.1 × 10⁴, 9 × 10³.

Method. Order by the power of 10 first; use the front number to break ties.

List the power of 10 of each number. The power of 10 is the main thing that decides the size.
2.5 × 10⁴: power 4
7.1 × 10⁻³: power −3
2.1 × 10⁴: power 4
9 × 10³: power 3
Order by the powers of 10. We need smallest to largest. Remember that −7 is smaller than −3.
power −3, power 3, power 4, power 4
If two numbers have the same power, compare their front numbers. With equal powers, the bigger front number is the bigger number.
power 4: 2.5 and 2.1 → smaller first is 2.1
Write the numbers in order.
7.1 × 10⁻³ , 9 × 10³ , 2.1 × 10⁴ , 2.5 × 10⁴

Check. As ordinary numbers: 0.0071, 9000, 21000, 25000 — these increase along the list. ✓

Watch out. A negative power is smaller than a positive power, and a larger negative power (like −8) is smaller than −2.

Worked example 6.5.3

Work out (2.5 × 10⁶) × (4 × 10⁻²). Give your answer in standard form.

Method. Multiply the front numbers, add the powers of 10, then make sure the front number is between 1 and 10.

Write the multiplication with the number parts together and the powers of 10 together. Multiplication can be done in any order, so we can group the front numbers and group the powers of 10.
(2.5 × 10⁶) × (4 × 10⁻²) = (2.5 × 4) × (10⁶ × 10⁻²)
Multiply the front numbers. Do this like an ordinary decimal multiplication.
Ignore the decimal points: 25 × 4 = 100
There are 1 decimal place altogether in 2.5 and 4, so the answer has 1 decimal place: 2.5 × 4 = 10
Multiply the powers of 10: add the indices. 10^a × 10^b = 10^a+b.
10⁶ × 10⁻² = 10^{6 + (−2)} = 10⁴
Put the two parts together.
10 × 10⁴
Check: is the front number between 1 and 10? 10 is 10 or more, so this is not standard form yet. Write 10 = 1 × 10 and combine the powers of 10.
10 × 10⁴ = 1 × 10 × 10⁴ = 1 × 10^{4 + 1} = 1 × 10⁵
Write the answer.
1 × 10⁵

Check. As ordinary numbers: 2500000 × 0.04 = 100000, which is 1 × 10⁵. ✓

Watch out. Add the powers when you multiply (do not multiply them). Always check the front number at the end: 12 × 10⁷ is not in standard form.

Worked example 6.5.4

Work out (3 × 10⁵) ÷ (5 × 10⁻²).

Method. Divide the front numbers, subtract the powers of 10, then make sure the front number is between 1 and 10.

Write the division with the number parts together and the powers of 10 together. We can split the division into the front numbers and the powers of 10.
(3 × 10⁵) ÷ (5 × 10⁻²) = (3 ÷ 5) × (10⁵ ÷ 10⁻²)
Divide the front numbers. Do this like an ordinary decimal division.
3 ÷ 5 = 0.6
Divide the powers of 10: subtract the indices. 10^a ÷ 10^b = 10^a−b.
10⁵ ÷ 10⁻² = 10^{5 − (−2)} = 10⁷
Put the two parts together.
0.6 × 10⁷
Check: is the front number between 1 and 10? 0.6 is less than 1, so this is not standard form yet. Write 0.6 = 6 × 10⁻¹ and combine the powers of 10.
0.6 × 10⁷ = 6 × 10⁻¹ × 10⁷ = 6 × 10^{−1 + 7} = 6 × 10⁶
Write the answer.
6 × 10⁶

Check. Multiply back: 6 × 5 = 30 and the powers give 10⁶ × 10⁻² = 10⁴, so we get 30 × 10⁴ = 300000, which is the number we started with. ✓

Watch out. Subtract the powers when you divide, and take care with negative powers: 4 − (−2) = 6. If the front number comes out less than 1, adjust it.

Worked example 6.5.5

Work out 3.2 × 10⁴ + 8.5 × 10³.

Method. Add standard form numbers: give them the same power of 10, add the front numbers, then tidy the answer back into standard form.

Make the powers of 10 the same. We can only add when the powers of 10 match, just as we can only add fractions with the same denominator. Change the number with the smaller power so that it uses 10⁴. Each increase of 1 in the power means moving the decimal point of the front number one place to the left.
8.5 × 10³ = 0.85 × 10⁴
Add the front numbers. The power of 10 stays the same: it just tells us the size of the unit.
3.2 + 0.85 = 4.05
so the result is 4.05 × 10⁴
Write the answer.
4.05 × 10⁴

Check. As ordinary numbers: 32000 + 8500 = 40500, which is 4.05 × 10⁴. ✓

Watch out. You cannot just add the front numbers when the powers are different: 3 × 10⁴ + 2 × 10³ is not 5 × 10⁷.

Worked example 6.5.6

Light travels at 3 × 10⁸ m/s. The Sun is 1.5 × 10¹¹ m from the Earth. How long does light take to reach the Earth? Give your answer in seconds in standard form.

Method. Divide the front numbers, subtract the powers of 10, then make sure the front number is between 1 and 10.

Decide what to do. The Sun is a long way from the Earth. We want a time. Time = distance ÷ speed, so divide the distance by the speed of light.
time = (1.5 × 10¹¹) ÷ (3 × 10⁸)
Write the division with the number parts together and the powers of 10 together. We can split the division into the front numbers and the powers of 10.
(1.5 × 10¹¹) ÷ (3 × 10⁸) = (1.5 ÷ 3) × (10¹¹ ÷ 10⁸)
Divide the front numbers. Do this like an ordinary decimal division.
Multiply both numbers by 10 to make them whole (the answer does not change): 1.5 ÷ 3 = 15 ÷ 30 = 0.5
Divide the powers of 10: subtract the indices. 10^a ÷ 10^b = 10^a−b.
10¹¹ ÷ 10⁸ = 10^{11 − 8} = 10³
Put the two parts together.
0.5 × 10³
Check: is the front number between 1 and 10? 0.5 is less than 1, so this is not standard form yet. Write 0.5 = 5 × 10⁻¹ and combine the powers of 10.
0.5 × 10³ = 5 × 10⁻¹ × 10³ = 5 × 10^{−1 + 3} = 5 × 10²
Write the answer with its unit.
5 × 10² s

Check. Multiply back: 5 × 3 = 15 and the powers give 10² × 10⁸ = 10¹⁰, so we get 15 × 10¹⁰ = 150000000000, which is the number we started with. ✓

Watch out. Subtract the powers when you divide, and take care with negative powers: 4 − (−2) = 6. If the front number comes out less than 1, adjust it.

Worked example 6.5.7

Use a calculator to work out (4.5 × 10⁷) × (2.4 × 10⁻³). Give your answer in standard form to 3 significant figures.

Method. Use the ×10^x key on a calculator, then write the display in standard form to the accuracy asked for.

Enter the calculation using the ×10^x key. The ×10^x key (on some calculators it is EXP) types "× 10 to the power" in one go. For a negative power, press the (−) key before the power number. Brackets stop the calculator mixing the numbers up.
Press: 4.5 [×10^x] 7 [×] 2.4 [×10^x] [(−)] 3 [=]
Read the answer from the display. Your calculator may show the answer as an ordinary number or in its own standard form style, such as 2.4E−3. The digits are what matter.
Display: 1.08 × 10⁵ (about)
Round to 3 significant figures and write in standard form. Round the front number A to 3 significant figures: look at the digit after the last one you keep, and round up if it is 5 or more. The power of 10 stays the same unless the rounding makes A reach 10.
1.08 × 10⁵

Check. Estimate by rounding each front number to the nearest whole number: (5 × 10⁷) × (2 × 10⁻³) ≈ 1 × 10⁵, which is close to our answer. ✓

Watch out. Use the ×10^x key, not separate × 10 ^ keys, unless you use brackets. Always check the power is sensible with a quick estimate.

Worked example 6.5.8

Work out (2 × 10³)³.

Method. Raise the front number to the power, multiply the index of 10 by the power, then tidy up.

Write the power of a product as separate powers. Both the front number and the power of 10 are raised to the power 3: (A × 10ⁿ)³ = A³ × (10ⁿ)³.
(2 × 10³)³ = 2³ × (10³)³
Work out 2³. Multiply 2 by itself 3 times.
2³ = 2 × 2 × 2 = 8
Work out the power of 10: multiply the indices. (10^a)^b = 10^a×b.
(10³)³ = 10^{3 × 3} = 10⁹
Put the parts together.
8 × 10⁹
Check the front number. It is between 1 and 10, so this is already in standard form.
8 × 10⁹

Check. As ordinary numbers: 2000³ = 8000000000, which is 8 × 10⁹. ✓

Watch out. The power applies to both parts. (2 × 10³)² is 4 × 10⁶, not 2 × 10⁶.

Worked example 6.5.9

Work out √(6.4 × 10⁷).

Method. Make the power of 10 even, take the square root of the front number, and halve the power.

Make the power of 10 even. We need to halve the index, and an odd index cannot be halved to a whole number. Move one factor of 10 into the front number.
6.4 × 10⁷ = 64 × 10⁶
Take the square root of each part. √(a × b) = √a × √b.
√64 = 8 (because 8 × 8 = 64) and √10⁶ = 10^{6 ÷ 2} = 10³
Put the parts together.
8 × 10³

Check. Square the answer: (8 × 10³)² = 64 × 10⁶, which is 6.4 × 10⁷ after tidying. ✓

Watch out. Halve the power of 10: √(10⁶) = 10³, not 10⁶ ÷ 2 = 5 × 10⁵.

Watch out

Do not add the powers when you add numbers: 3 × 10⁴ + 2 × 10³ is not 5 × 10⁷. Make the powers match first. And after multiplying, check A: 12 × 10⁷ is not in standard form.

Exam tip

In calculator questions write down the calculation you typed, then the answer, then round. If a question says “in standard form” a plain decimal answer will lose marks.

Exercise 6K Comparing and ordering standard form

Non-calculator

FWrite down the larger number in each pair.

(a)3.2 × 10⁵ and 9.9 × 10⁴

(b)4.5 × 10⁻³ and 4.5 × 10⁻⁵

(d)2 × 10⁻⁶ and 9 × 10⁻⁷

(e)1.2 × 10¹⁰ and 9.9 × 10⁹

Done

FPut < or > between each pair.

(a)7.5 × 10³ ? 2.1 × 10⁶

(b)3.3 × 10⁻² ? 1.9 × 10⁻⁴

(d)9 × 10⁻⁵ ? 1 × 10⁻⁴

Done

FWrite each list in order, smallest first.

(a)2.5 × 10⁴ , 7.1 × 10⁻³ , 2.1 × 10⁴ , 9 × 10³

(b)5.2 × 10⁻⁶ , 4.9 × 10⁻⁵ , 1.1 × 10⁻⁵ , 8 × 10⁻⁷

Done

FWrite this list in order, largest first: 6 × 10⁸, 6 × 10⁻⁸, 5.9 × 10⁸, 1 × 10⁹.

Done

PSThe distances of four planets from the Sun, in km, are: Mercury 5.79 × 10⁷, Venus 1.08 × 10⁸, Earth 1.5 × 10⁸, Mars 2.28 × 10⁸. Write the planets in order, closest to the Sun first.

Done

PSThe mass of an electron is 9.11 × 10⁻³¹ kg. The mass of a proton is 1.67 × 10⁻²⁷ kg. Which has the smaller mass?

Done

RExplain why 9.9 × 10² is less than 1.1 × 10⁵, even though 9.9 is bigger than 1.1.

Done

RWrite down a number in standard form that is bigger than 3 × 10⁴ and smaller than 3 × 10⁵.

Done

RTrue or false: 5 × 10⁻³ > 5 × 10⁻²? Explain.

Done

6.6

Exact answers with π

Edexcel 1MA1: N8, G17
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Exact answers with πComing soon

The number π (pi) is the circumference of any circle divided by its diameter. Its decimal never ends and never repeats: 3.14159265…. Because it cannot be written exactly as a decimal or fraction, any answer found with a rounded value of π is only approximate. If you want an exact answer you leave π as the symbol π. Questions say “give your answer in terms of π” or “leave your answer as a multiple of π”.

The area of a circle is π × r × r. For radius 6 cm the exact area is π × 36 = 36π cm².

Remember the formulae: circumference C = πd = 2πr and area A = πr² (more on circles in Chapter 18). Square only the radius, then write the number first and π after it: 36π, not π36. Treat π like a letter in algebra. 3π + 5π = 8π (like 3x + 5x = 8x), 4 × 3π = 12π, and 12π ÷ 4 = 3π.

Parts of circles use fractions: a semicircle is half and a quarter circle is one quarter. Take the fraction of the exact answer: a quarter of 64π is 16π. For a perimeter, the curved part is a fraction of the circumference and the straight edges are ordinary lengths, so the answer has two terms, such as (5π + 10) cm. Compound shapes: find each area exactly and add. If you need a decimal at the end, use the π button on your calculator and round at the very end.

A rectangle 8 cm by 5 cm with a semicircle of diameter 8 cm: area = 40 + 8π cm².

Key ideas

Exact answer: leave π in the answer, with the number in front: 36π.
C = πd = 2πr and A = πr². Use the radius for the area (half the diameter).
Treat π like a letter: 3π + 5π = 8π.
A fraction of a circle: take that fraction of the exact area (or circumference).
Only use the π button, and round, when you want a decimal.

Worked example 6.6.1

Work out the area of a circle with radius 6 cm. Give your answer in terms of π.

Method. Put the numbers into the formula but leave π as the symbol π. Then the answer is exact.

Write the formula for the area of a circle. The area of a circle is π times the radius squared.
A = πr²
Substitute the radius. Replace r with the radius. Leave π as the symbol.
A = π × 6²
Work out the square first. In BIDMAS, powers come before multiplication. Square the radius, not the π.
6² = 6 × 6 = 36
Write the number in front of π. Write the exact answer with the number first and then π, and add the units.
36π cm²

Check. Use the π button to get a decimal and compare: 36π = 113.097 to 3 decimal places, so about 113.1 cm² to 1 d.p. This is similar in size to 113.04 found using 3.14. ✓

Watch out. Do not replace π by 3.14 or press the π key if the question says "in terms of π" or "give your answer exactly". Also square only the radius.

Worked example 6.6.2

Work out the circumference of a circle with diameter 10 cm, in terms of π.

Method. Put the numbers into the formula but leave π as the symbol π. Then the answer is exact.

Find the radius. The radius is half of the diameter.
radius = 10 ÷ 2 = 5 cm
Write the formula for the circumference. The circumference is the distance around the circle: π times the diameter, which is also 2 × π × radius.
C = πd or C = 2πr
Substitute. Use the diameter, which is double the radius. Leave π as the symbol.
d = 2 × 5 = 10 cm
C = π × 10
Write the number in front of π. Write the exact answer with the number first and then π, and add the units.
10π cm

Check. Use the π button to get a decimal and compare: 10π = 31.416 to 3 decimal places, so about 31.4 cm to 1 d.p. This is similar in size to 31.4 found using 3.14. ✓

Watch out. Do not replace π by 3.14 or press the π key if the question says "in terms of π" or "give your answer exactly". Also square only the radius.

Worked example 6.6.3

Work out 3π + 5π.

Method. Treat π as if it were a letter. Work with the numbers in front of it and keep π in the answer.

Treat π like a letter in algebra. It is a number, but we do not write it as a decimal. So 3π + 5π works just like 3x + 5x = 8x: 3 lots of π and 5 lots of π make 8 lots of π.
3π + 5π
Add the numbers in front of π. The number in front tells us how many lots of π there are.
3 + 5 = 8
Write the answer with π.
8π

Check. Use 3.14 for π as a rough check: 3 lots of 3.14 and the answer 8 lots of 3.14 = 25.12, which has the right size. ✓

Watch out. π does not change when you add or subtract numbers in front of it. 3π + 5π = 8π, not 8π².

Worked example 6.6.4

The area of a quarter circle of radius 8 cm. Leave π in your answer.

Method. Find the area of the whole circle, then take the fraction of it that the shape is. Leave π in the answer.

Write the area of the whole circle. A = πr².
A = π × 8² = 64π cm²
Take the fraction of it. The shape is a quarter of the circle, so multiply by 14.
14 × 64π = 16π
Write the answer with units.
16π cm²

Check. As a decimal this is about 50.3 cm², which is less than the whole circle (201.1 cm²). ✓

Watch out. Use the radius. If you are given the diameter, halve it first.

Worked example 6.6.5

Work out the perimeter of a semicircle of radius 5 cm. Leave your answer in terms of π.

Method. Add the curved length (a fraction of the circumference) to the straight edges. Leave π in the answer.

Split the perimeter into the curved part and the straight part(s). The perimeter is the distance all the way round the shape. A semicircle has one curved edge and straight edges.
perimeter = curved edge + straight edge(s)
Find the length of the curved edge. The curved edge is half of the circumference of the full circle.
half of 2πr = πr = π × 5 = 5π cm
Find the length of the straight edge(s).
the diameter = 2 × 5 = 10 cm
Add them. Leave π in the answer. A multiple of π and a whole number cannot be combined into one term, so the answer has two parts.
(5π + 10) cm

Check. As a decimal: 15.71 + 10 = 25.71 cm (about). This is a sensible length for a shape with radius 5 cm. ✓

Watch out. Do not forget the straight edge(s). The perimeter of a semicircle is more than half the circumference of the circle.

Worked example 6.6.6

Work out the exact area of a rectangle 8 cm by 5 cm with a semicircle of diameter 8 cm on one end.

Method. Split into a rectangle and a semicircle, find each area exactly, then add.

Split the shape into a rectangle and a semicircle. It is easier to find the areas of simple shapes and add them.
area = rectangle + semicircle
Find the area of the rectangle. area = length × width
8 × 5 = 40 cm²
Find the radius of the semicircle. The diameter of the semicircle is the width of the rectangle. The radius is half the diameter.
radius = 8 ÷ 2 = 4 cm
Find the area of the semicircle. A semicircle is half a circle: half of πr².
12 × π × 4² = 12 × π × 16 = 8π cm²
Add the two areas. Leave π in the answer. The whole numbers and the π part cannot be combined into one term.
(40 + 8π) cm²

Check. As a decimal: 40 + 25.13 = 65.13 cm² (about). The rectangle alone is 40 cm², so this is a little bigger. ✓

Watch out. Use the radius (half the width) in the circle formula, not the whole width.

Worked example 6.6.7

The area of a circle is 20π cm². Find its radius as a simplified surd.

Key words. π (pi) is the number 3.14159… that you get when you divide the circumference of any circle by its diameter. It never ends and never repeats, so we cannot write it exactly as a decimal. An answer left in terms of π (such as 12π) is exact: nothing has been rounded. Key words. A surd is a square root that cannot be written as a whole number or fraction, such as √2 or √7 (√9 = 3 is not a surd). Surds are irrational: their decimals never end or repeat. We keep them as roots to give exact answers. The two rules to remember are √a × √b = √(ab) and √a × √a = a.

Method. Use the area formula backwards: divide by π, take the square root, and simplify the surd.

Write the formula for the area.
A = πr²
Put in the area and divide both sides by π. The area is 20π cm². Dividing both sides by π leaves the number.
20π = πr², so r² = 20
Take the square root and leave it as a surd. 20 is not a perfect square, so keep it exact and simplify the surd: the largest square factor of 20 is 4.
r = √20 = √(4 × 5) = 2√5
Write the answer.
2√5 cm

Check. Put it back: π × (2√5)² = π × 20 = 20π. ✓

Watch out. Leave the radius as a simplified surd for an exact answer.

Watch out

Do not replace π by 3.14 in an exact answer, and do not square the π. Area is π × (radius)²: for radius 6 it is 36π, not 6π. And use the radius: if you are given the diameter, halve it first.

6.7

Surds: simplifying and adding

Edexcel 1MA1: N8, A4
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Surds: simplifying and addingComing soon

Some square roots are whole numbers: √49 = 7. Most are not. √2 = 1.41421356… never ends and never repeats, so no fraction or terminating decimal equals it exactly. A root like this is called a surd (√2, √3, √5, √7…; but √9 = 3 is not a surd). Surds are irrational numbers. To give an exact answer we leave them as roots and keep them as simple as possible. Any decimal for a surd is only an approximation.

Surds are real lengths: a right-angled triangle with two short sides 1 has hypotenuse √2. Adding more triangles gives √3, √4 = 2, √5, …

Two rules do almost all the work: √a × √b = √(ab), and √a × √a = a. Used backwards, the first rule lets us simplify a surd. √12 = √(4 × 3) = √4 × √3 = 2√3. To simplify √n: find the largest square number (4, 9, 16, 25, 36, 49, …) that divides into n, write n as (square) × (other), split the root, and take the root of the square number. A surd is fully simplified when the number under the root has no square factor. If there is a number in front, multiply it by the number you take out: 3√50 = 3 × 5√2 = 15√2.

Adding and subtracting surds works like collecting like terms. 3√2 + 5√2 = 8√2 (like 3 apples + 5 apples), but √2 + √3 cannot be combined. So simplify each surd first; they may then have the same number under the root. √12 + √27 = 2√3 + 3√3 = 5√3.

Key ideas

A surd is a square root that is not a whole number (or fraction), such as √7. It is irrational.
√a × √b = √(ab) and √a × √a = a.
To simplify √n: use the largest square factor. √12 = √(4 × 3) = 2√3.
Add and subtract only like surds (same number under the root). Simplify each surd first.

Worked example 6.7.1

Simplify √12.

Key words. A surd is a square root that cannot be written as a whole number or fraction, such as √2 or √7 (√9 = 3 is not a surd). Surds are irrational: their decimals never end or repeat. We keep them as roots to give exact answers. The two rules to remember are √a × √b = √(ab) and √a × √a = a.

Method. Find the largest square number that is a factor of the number under the root. Split the root into a product, and take the root of the square number.

List the square numbers and see which ones divide into 12. To simplify a surd we look for a square number factor: 4, 9, 16, 25, 36, 49, 64, 81, 100… The biggest one that divides exactly is the best one to use.
4: 12 ÷ 4 = 3 ✓ goes in exactly
9: 12 ÷ 9 = 1 remainder 3 ✗ does not go in exactly
Write 12 as 4 × 3. 4 is the largest square number that divides into 12. Use it so that the answer is fully simplified.
12 = 4 × 3, so √12 = √4 × 3
Split the root into two roots. The rule is √(a × b) = √a × √b. This works because squaring each side gives the same number.
√4 × 3 = √4 × √3
Work out √4. 4 is a square number, so its root is a whole number: 2 × 2 = 4.
√4 = 2, so √4 × √3 = 2√3
Write the answer.
2√3

Check. Square the answer: (2√3)² = 2² × 3 = 4 × 3 = 12, and the original squared is (√12)² = 12. ✓

Watch out. If you choose a smaller square factor you must carry on simplifying. For example √72 = √(4 × 18) = 2√18, and √18 simplifies again to 3√2. Always check whether the surd left can be simplified more.

Worked example 6.7.2

Simplify 3√50.

Method. Find the largest square number that is a factor of the number under the root. Split the root into a product, and take the root of the square number.

List the square numbers and see which ones divide into 50. To simplify a surd we look for a square number factor: 4, 9, 16, 25, 36, 49, 64, 81, 100… The biggest one that divides exactly is the best one to use.
4: 50 ÷ 4 = 12 remainder 2 ✗ does not go in exactly
9: 50 ÷ 9 = 5 remainder 5 ✗ does not go in exactly
16: 50 ÷ 16 = 3 remainder 2 ✗ does not go in exactly
25: 50 ÷ 25 = 2 ✓ goes in exactly
36: 50 ÷ 36 = 1 remainder 14 ✗ does not go in exactly
49: 50 ÷ 49 = 1 remainder 1 ✗ does not go in exactly
Write 50 as 25 × 2. 25 is the largest square number that divides into 50. Use it so that the answer is fully simplified.
50 = 25 × 2, so √50 = √25 × 2
Split the root into two roots. The rule is √(a × b) = √a × √b. This works because squaring each side gives the same number.
√25 × 2 = √25 × √2
Work out √25. 25 is a square number, so its root is a whole number: 5 × 5 = 25.
√25 = 5, so √25 × √2 = 5√2
Multiply by the number that was already in front. The original expression was 3 lots of √50, so multiply the 5 by 3.
3 × 5√2 = 15√2
Write the answer.
15√2

Check. Square the answer: (15√2)² = 15² × 2 = 225 × 2 = 450, and the original squared is (3√50)² = 450. ✓

Worked example 6.7.3

Simplify √35 if possible.

Method. Look for the largest square number that divides into the number under the root.

List the square numbers and see which ones divide into 35. To simplify a surd we look for a square number factor: 4, 9, 16, 25, 36, 49, 64, 81, 100… The biggest one that divides exactly is the best one to use.
4: 35 ÷ 4 = 8 remainder 3 ✗ does not go in exactly
9: 35 ÷ 9 = 3 remainder 8 ✗ does not go in exactly
16: 35 ÷ 16 = 2 remainder 3 ✗ does not go in exactly
25: 35 ÷ 25 = 1 remainder 10 ✗ does not go in exactly
Decide. No square number except 1 divides into 35, so √35 is already in its simplest form.
√35 cannot be simplified.
Write the answer. A surd is in its simplest form when the number under the root has no square factor (other than 1).
√35

Check. Check: no square number between 2 and 35 divides exactly into 35. ✓

Worked example 6.7.4

Simplify √12 + √27.

Method. Simplify every surd first. Then add or subtract the like surds (same number under the root).

Simplify each surd on its own. Surds can only be added or subtracted when the number under the root is the same (like terms). Simplifying each surd often makes them match.
√12 = √4 × 3 = 2√3
√27 = √9 × 3 = 3√3
Rewrite the expression with the simplified surds. Keep the signs the same as before.
2√3 + 3√3
Collect the like surds. Like surds have the same number under the root. Add the numbers in front, just as you would add 3 apples and 5 apples. The root stays the same.
√3 terms: 2 + 3 = 5, so 5√3
Write the answer.
5√3

Check. Use decimals: the original is about 8.660 and the answer is about 8.660. They agree. ✓

Watch out. Only add or subtract surds that have the same number under the root. √2 + √3 is not √5.

Worked example 6.7.5

Simplify 5√2 − √8 + 2√18.

Method. Simplify every surd first. Then add or subtract the like surds (same number under the root).

Simplify each surd on its own. Surds can only be added or subtracted when the number under the root is the same (like terms). Simplifying each surd often makes them match.
5√2: already simplest
√8 = √4 × 2 = 2√2
2√18 = 2 × √9 × 2 = 2 × 3√2 = 6√2
Rewrite the expression with the simplified surds. Keep the signs the same as before.
5√2 − 2√2 + 6√2
Collect the like surds. Like surds have the same number under the root. Add the numbers in front, just as you would add 3 apples and 5 apples. The root stays the same.
√2 terms: 5 − 2 + 6 = 9, so 9√2
Write the answer.
9√2

Check. Use decimals: the original is about 12.728 and the answer is about 12.728. They agree. ✓

Watch out. Only add or subtract surds that have the same number under the root. √2 + √3 is not √5.

Worked example 6.7.6

Write √(2/9) in its simplest form.

Method. Take the square root of the top and the bottom separately, then simplify each.

Split the root of a fraction. √(a/b) = √a ÷ √b.
√29 = √2√9
Simplify the top and the bottom. Take square factors out of each root.
√2 = √2, √9 = 3
Write the answer as a fraction. Cancel any common factor.
√23

Check. Decimals: √(2/9) = 0.4714 and the answer is 0.4714. ✓

Watch out. Only the bottom can have its surd removed later by rationalising (section 6.9).

Watch out

√2 + √3 is not √5, and √(9 + 16) is not √9 + √16. You can only add surds with the same number under the root. Also use the largest square factor, or check afterwards that nothing under the root can still be simplified.

6.8

Multiplying, dividing and expanding with surds

Edexcel 1MA1: N8, A4
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Multiplying, dividing and expanding with surdsComing soon

To multiply surds, multiply the numbers in front and multiply the numbers under the roots: (2√3) × (5√6) = 10√18 = 10 × 3√2 = 30√2. Always simplify at the end. When a surd is multiplied by itself you get a whole number: √3 × √3 = 3, and (2√3)² = 4 × 3 = 12. To divide, divide the numbers in front and the numbers under the roots: 6√10 ÷ 2√5 = 3√2.

Expanding brackets with surds uses exactly the same method as in algebra: multiply every term in one bracket by every term in the other. The picture shows (2 + √3)(1 + √3) as the area of a rectangle in four parts.

The four parts are 2, 2√3, √3 and 3. Adding gives 5 + 3√3.

Remember the special products. (a + √b)² = a² + 2a√b + b: do not forget the middle term. And the difference of two squares: (a + √b)(a − √b) = a² − b. The two middle terms cancel, and the surd disappears. This last result is the key to rationalising the denominator in section 6.9.

Key ideas

(a√b) × (c√d) = ac√(bd); then simplify.
√a × √a = a; (c√a)² = c²a.
Expand brackets: every term × every term (First, Outer, Inner, Last), then collect like terms.
(a + √b)(a − √b) = a² − b: a whole number.

Worked example 6.8.1

Simplify (2√3) × (5√6).

Method. Multiply the numbers in front, multiply the numbers under the roots, then simplify.

Multiply the numbers outside the roots. The numbers in front of the roots are multiplied together.
2 × 5 = 10
Multiply the numbers inside the roots. The rule is √a × √b = √(ab).
√3 × √6 = √18
Put them together.
10√18
Simplify the surd if you can. Look for a square number that divides into 18.
√18 = √9 × 2 = 3√2
10 × 3√2 = 30√2
Write the answer.
30√2

Check. Decimals: 3.464 × 12.247 ≈ 42.43, and the answer is about 42.43. ✓

Watch out. Do not multiply the number in front with the number under the root. And √3 × √3 = 3 (a whole number), not √9 × 2.

Worked example 6.8.2

Simplify 6√10 ÷ 2√5.

Method. Divide the numbers in front, divide the numbers under the roots, then simplify.

Divide the numbers outside the roots. The numbers in front of the roots are divided.
6 ÷ 2 = 3
Divide the numbers inside the roots. The rule is √a ÷ √b = √(a ÷ b).
√10 ÷ √5 = √10 ÷ 5 = √2
Put them together and simplify. If the number under the root is a square number, take its root.
3√2

Check. Check by multiplying back: 3√2 × 2√5 = 6√10, which is 6√10 (simplified). ✓

Watch out. Divide the numbers under the roots, not the roots themselves.

Worked example 6.8.3

Expand and simplify (3 + √2)(4 − √2).

Method. Expand the brackets by multiplying every term by every term. Simplify each product, then collect like terms.

Write the product. We multiply every term in the first bracket by every term in the second bracket. A way to remember the four products is First, Outer, Inner, Last (FOIL).
(3 + √2)(4 − √2)
Multiply term by term. Simplify each product. Use √a × √a = a and √a × √b = √(ab). Take care with the signs.
First: 3 × 4 = 12
Outer: 3 × (−√2) = −3√2
Inner: √2 × 4 = 4√2
Last: √2 × (−√2) = −√4 = −2
Write all the simplified terms in one line. Keep the signs.
12 − 3√2 + 4√2 − 2
Collect like terms. Add the whole numbers together. Add the like surds together (the same number under the root).
10 + √2

Check. Decimals: the original is 4.414 × 2.586 ≈ 11.41 and the answer is about 11.41. ✓

Watch out. (2 + √3)² is not 4 + 3. Always write out both brackets: (2 + √3)(2 + √3). The middle terms do not disappear.

Worked example 6.8.4

Expand and simplify (2 + √3)².

Method. Expand the brackets by multiplying every term by every term. Simplify each product, then collect like terms.

Write the product. We multiply every term in the first bracket by every term in the second bracket. A way to remember the four products is First, Outer, Inner, Last (FOIL).
(2 + √3)(2 + √3)
Multiply term by term. Simplify each product. Use √a × √a = a and √a × √b = √(ab). Take care with the signs.
First: 2 × 2 = 4
Outer: 2 × √3 = 2√3
Inner: √3 × 2 = 2√3
Last: √3 × √3 = √9 = 3
Write all the simplified terms in one line. Keep the signs.
4 + 2√3 + 2√3 + 3
Collect like terms. Add the whole numbers together. Add the like surds together (the same number under the root).
7 + 4√3

Check. Decimals: the original is 3.732 × 3.732 ≈ 13.93 and the answer is about 13.93. ✓

Watch out. (2 + √3)² is not 4 + 3. Always write out both brackets: (2 + √3)(2 + √3). The middle terms do not disappear.

Worked example 6.8.5

Expand and simplify (4 + √5)(4 − √5).

Method. Expand the brackets by multiplying every term by every term. Simplify each product, then collect like terms.

Write the product. We multiply every term in the first bracket by every term in the second bracket. A way to remember the four products is First, Outer, Inner, Last (FOIL).
(4 + √5)(4 − √5)
Multiply term by term. Simplify each product. Use √a × √a = a and √a × √b = √(ab). Take care with the signs.
First: 4 × 4 = 16
Outer: 4 × (−√5) = −4√5
Inner: √5 × 4 = 4√5
Last: √5 × (−√5) = −√25 = −5
Write all the simplified terms in one line. Keep the signs.
16 − 4√5 + 4√5 − 5
Collect like terms. Add the whole numbers together. Add the like surds together (the same number under the root).
11

Check. Decimals: the original is 6.236 × 1.764 ≈ 11.00 and the answer is about 11.00. ✓

Watch out. (2 + √3)² is not 4 + 3. Always write out both brackets: (2 + √3)(2 + √3). The middle terms do not disappear.

Worked example 6.8.6

Expand and simplify √6(√2 + √3).

Method. Expand the brackets by multiplying every term by every term. Simplify each product, then collect like terms.

Write the product. We multiply every term in the first bracket by every term in the second bracket.
√6(√2 + √3)
Multiply term by term. Simplify each product. Use √a × √a = a and √a × √b = √(ab). Take care with the signs.
√6 × √2 = √12 = √4 × 3 = 2√3
√6 × √3 = √18 = √9 × 2 = 3√2
Write all the simplified terms in one line. Keep the signs.
2√3 + 3√2
Collect like terms. Add the whole numbers together. Add the like surds together (the same number under the root).
3√2 + 2√3

Check. Decimals: the original is 2.449 × 3.146 ≈ 7.71 and the answer is about 7.71. ✓

Watch out. (2 + √3)² is not 4 + 3. Always write out both brackets: (2 + √3)(2 + √3). The middle terms do not disappear.

Watch out

(2 + √3)² is not 4 + 3. Write the two brackets out: (2 + √3)(2 + √3) = 4 + 2√3 + 2√3 + 3 = 7 + 4√3. Do not lose the middle terms.

6.9

Rationalising the denominator and exact answers

Edexcel 1MA1: N8, A4
Grade 8-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Rationalising the denominator and exact answersComing soon

A fraction with a surd in its denominator, such as 1/√2, is awkward to work with, and it is not considered fully simplified. Rationalising the denominator means rewriting the fraction so that the bottom is a whole number. We use the fact that multiplying the top and the bottom of a fraction by the same number does not change its value, and that √2 × √2 = 2:

1√2 = 1√2 × √2√2 = √22

A right-angled isosceles triangle with hypotenuse 1 has short sides 1/√2. Rationalised, that is √2/2 ≈ 0.707, which is easier to picture and to calculate with.

Type 1: a single surd on the bottom (such as 6/√3 or 3/(2√5)). Multiply the top and bottom by that surd, then simplify: 6/√3 = 6√3/3 = 2√3.

Type 2: a whole number plus or minus a surd (such as 1/(3 + √5)). Multiplying by the surd alone does not work, because the bottom has two terms. Instead multiply top and bottom by the conjugate: the same two terms with the sign in the middle changed. The conjugate of 3 + √5 is 3 − √5. By the difference of two squares, (3 + √5)(3 − √5) = 9 − 5 = 4, a whole number. So 1/(3 + √5) = (3 − √5)/4.

Surds also give exact answers in geometry. In Pythagoras, if c² = 52 then c = √52 = 2√13 cm exactly. Give exact answers unless the question asks you to round.

Key ideas

Rationalise: multiply top and bottom by the same thing so the bottom becomes a whole number.
For k/(c√n) multiply top and bottom by √n.
For k/(a + √b) multiply by the conjugate (a − √b); the bottom becomes a² − b.
Always simplify the final fraction (divide the top and bottom by common factors).

Worked example 6.9.1

Rationalise the denominator of 6√3.

Method. Multiply the top and bottom by the surd in the denominator. This makes the denominator a whole number.

Say what is wrong with the fraction. A denominator (the bottom) containing a surd is not allowed in a fully simplified answer. Rationalising means rewriting the fraction so that the bottom is a whole number.
6√3
Multiply the top and the bottom by √3. √3 ÷ √3 = 1, so multiplying by it does not change the value of the fraction. We choose √3 because √3 × √3 = 3, a whole number.
6√3 × √3√3 = 6√3√3×√3
Work out the top and the bottom. Top: multiply each term by √3. Bottom: √3 × √3 = 3.
6√33
Simplify the fraction. Divide the top and the bottom by any number that goes into every term.
2√3

Check. Decimals: 3.464 before and 3.464 after. They agree. ✓

Watch out. You must multiply the top and the bottom. If you only multiply the bottom, you change the value of the fraction.

Worked example 6.9.2

Rationalise the denominator of 32√5.

Method. Multiply the top and bottom by the surd in the denominator. This makes the denominator a whole number.

Say what is wrong with the fraction. A denominator (the bottom) containing a surd is not allowed in a fully simplified answer. Rationalising means rewriting the fraction so that the bottom is a whole number.
32√5
Multiply the top and the bottom by √5. √5 ÷ √5 = 1, so multiplying by it does not change the value of the fraction. We choose √5 because √5 × √5 = 5, a whole number.
32√5 × √5√5 = 3√52×√5×√5
Work out the top and the bottom. Top: multiply each term by √5. Bottom: √5 × √5 = 5.
3√510
The fraction cannot be simplified further.
3√510

Check. Decimals: 0.671 before and 0.671 after. They agree. ✓

Watch out. You must multiply the top and the bottom. If you only multiply the bottom, you change the value of the fraction.

Worked example 6.9.3

Write 1 + √3√2 with a rational denominator.

Method. Multiply the top and bottom by the surd in the denominator. This makes the denominator a whole number.

Say what is wrong with the fraction. A denominator (the bottom) containing a surd is not allowed in a fully simplified answer. Rationalising means rewriting the fraction so that the bottom is a whole number.
1 + √3√2
Multiply the top and the bottom by √2. √2 ÷ √2 = 1, so multiplying by it does not change the value of the fraction. We choose √2 because √2 × √2 = 2, a whole number.
1 + √3√2 × √2√2 = (1 + √3)√2√2×√2
Work out the top and the bottom. Top: multiply each term by √2. Bottom: √2 × √2 = 2.
√2 + √62
Top: (1 + √3)√2 = √2 + √6
The fraction cannot be simplified further.
√2 + √62

Check. Decimals: 1.932 before and 1.932 after. They agree. ✓

Watch out. You must multiply the top and the bottom. If you only multiply the bottom, you change the value of the fraction.

Worked example 6.9.4

Rationalise the denominator of 13 + √5.

Method. Multiply the top and bottom by the conjugate of the bottom. The bottom becomes a whole number (difference of two squares).

Say what is wrong with the fraction. The bottom contains a surd. We need to rationalise it (make the bottom a whole number). Multiplying by the surd alone does not work here, because the bottom has two terms.
13 + √5
Write down the conjugate of the bottom. The conjugate is the same two terms with the sign between them changed. Multiplying a bracket by its conjugate gives a difference of two squares, so the surd disappears.
bottom: 3 + √5 conjugate: 3 − √5
Multiply top and bottom by the conjugate. Multiplying the top and the bottom by the same thing is multiplying by 1, so the value does not change.
1(3 + √5) × (3 − √5)(3 − √5)
Expand the bottom using the difference of two squares. (p + q)(p − q) = p² − q², and the middle terms cancel.
(3 + √5)(3 − √5) = 3² − (√5)² = 9 − 5 = 4
Expand the top. Multiply the top by the conjugate, term by term.
1 × (3 − √5) = 3 − √5
Write the new fraction and simplify. Divide every term on the top by the number on the bottom if you can.
3 − √54 = 3 − √54

Check. Decimals: 0.191 before and 0.191 after. They agree. ✓

Watch out. Change the sign between the two terms to get the conjugate, and multiply both the top and the bottom by it.

Worked example 6.9.5

Write 64 − √2 in the form a + b√2c.

Method. Multiply the top and bottom by the conjugate of the bottom. The bottom becomes a whole number (difference of two squares).

Say what is wrong with the fraction. The bottom contains a surd. We need to rationalise it (make the bottom a whole number). Multiplying by the surd alone does not work here, because the bottom has two terms.
64 − √2
Write down the conjugate of the bottom. The conjugate is the same two terms with the sign between them changed. Multiplying a bracket by its conjugate gives a difference of two squares, so the surd disappears.
bottom: 4 − √2 conjugate: 4 + √2
Multiply top and bottom by the conjugate. Multiplying the top and the bottom by the same thing is multiplying by 1, so the value does not change.
6(4 − √2) × (4 + √2)(4 + √2)
Expand the bottom using the difference of two squares. (p + q)(p − q) = p² − q², and the middle terms cancel.
(4 − √2)(4 + √2) = 4² − (√2)² = 16 − 2 = 14
Expand the top. Multiply the top by the conjugate, term by term.
6 × (4 + √2) = 24 + 6√2
Write the new fraction and simplify. Divide every term on the top by the number on the bottom if you can.
24 + 6√214 = 12 + 3√27

Check. Decimals: 2.320 before and 2.320 after. They agree. ✓

Watch out. Change the sign between the two terms to get the conjugate, and multiply both the top and the bottom by it.

Worked example 6.9.6

Work out the exact length of the hypotenuse of a right-angled triangle with shorter sides 4 cm and 6 cm.

Method. Use Pythagoras to get the square of the length, then take the square root and leave it as a simplified surd.

Write Pythagoras' theorem. In a right-angled triangle the square on the longest side (the hypotenuse) equals the sum of the squares on the other two sides.
c² = a² + b²
Substitute the two shorter sides.
c² = 4² + 6² = 16 + 36 = 52
Take the square root. Leave it as a surd, so that the answer is exact.
side = √52
Simplify the surd. The largest square factor of 52 is 4.
√52 = √4 × 13 = 2√13
Write the answer with units. This is the exact length.
2√13 cm

Check. Square the answer: (2√13)² = 52. As a decimal the length is about 7.21 cm. ✓

Watch out. Do not round if the question asks for an exact answer. Simplify the surd.

Watch out

Multiply the top and the bottom. For a bottom like 3 + √5, use the conjugate 3 − √5, not just √5. Check that the surd really disappears from the bottom, and simplify the fraction at the end.

Exercise 6Q Rationalising the denominator

Non-calculator

FRationalise the denominator of each fraction. Simplify your answers.

(a)6√3

(b)1√7

(d)42√3

(e)12√6

(f)3√2

Done

FRationalise the denominator of each fraction. Simplify your answers.

(a)1 + √3√2

(b)2 + √5√5

(c)√2 + √6√2

(d)3 + √5√5

Done

FRationalise the denominator of each fraction. Simplify your answers.

(a)13 + √5

(b)64 − √2

(d)12 − √3

(e)53 + √2

Done

FRationalise the denominator of each fraction. Simplify your answers.

(a)2 + √31 + √3

(b)1 + √52 − √5

Done

FWrite each fraction in the form a + b√c, where a, b and c are integers.

(a)12 + √3

(b)43 − √5

Done

RShow that 1√3 − 1 = √3 + 12.

Done

RChloe multiplies only the bottom of 1√5 by √5 and writes 15. Explain her mistake and rationalise the fraction correctly.

Done

RExplain why you can multiply the top and the bottom of a fraction by the same number without changing its value.

Done

Exercise 6R Exact answers with surds

Non-calculator

PSWork out the exact length of the side marked x. Simplify your answers.

(a)shorter sides 4 cm and 6 cm; find the hypotenuse

(b)shorter sides 5 cm and 5 cm; find the hypotenuse

(c)shorter sides 1 cm and 3 cm; find the hypotenuse

(d)hypotenuse 7 cm and one side 3 cm; find the other side

Done

PSA square has sides of length 5 cm. Work out the exact length of its diagonal.

Done

PSAn equilateral triangle has sides 2 cm. Its height splits it into two right-angled triangles. Work out the exact height, and then the exact area of the equilateral triangle.

(a)height

(b)area

Done

RA triangle has a base of √12 cm and a perpendicular height of √3 cm. Show that its area is 3 cm².

Done

RA right-angled triangle has shorter sides √6 cm and √10 cm. Show that its hypotenuse is exactly 4 cm.

Done

RShow that each of these is equal to a multiple of √5.

(a)√45 + √20

(b)3√20 − √45

(c)√80 ÷ 2 + √5

Done

RGive an example to show that the product of two irrational numbers can be rational.

Done

RThe area of a circle is 12π cm². Show that its radius is 2√3 cm and find its exact circumference.

(a)radius

(b)circumference

Done

RShow that 1√2 + 1 + 1√2 − 1 = 2√2.

Done

Chapter summary

Skill	Method
Index laws (same base)	Multiply: add indices. Divide: subtract indices. Power of a power: multiply indices. a¹ = a.
Zero and negative indices	a⁰ = 1. a⁻ⁿ = 1/aⁿ. For fractions, flip the fraction and make the index positive.
Indices with letters	Deal with the number and each letter separately. (2a³)⁴ = 16a¹².
Roots	√a: number that multiplies by itself to give a. Between two whole numbers: use the perfect squares or cubes on either side. x² = k has two solutions.
Fractional indices (Higher)	a^m/n = (n-th root of a)^m. Do the root first. Negative: reciprocal first. Solve x^p/q = k by raising to the power q/p.
Standard form	A × 10ⁿ with 1 ≤ A < 10. Big number: positive n. Number less than 1: negative n. Count the places the decimal point moves.
Calculating in standard form	Multiply: multiply the A values, add the powers. Divide: divide the A values, subtract the powers. Add or subtract: make the powers match. Then tidy so that 1 ≤ A < 10.
Comparing standard form	Compare the powers of 10 first; if equal compare the front numbers.
Exact answers with π	Leave π in the answer: area 36π. Treat π like a letter. Use the radius for area.
Simplifying surds (Higher)	√a × √b = √(ab). Use the largest square factor: √12 = 2√3. Add or subtract only like surds.
Expanding with surds (Higher)	Every term × every term. √a × √a = a. (a + √b)(a − √b) = a² − b.
Rationalising (Higher)	k/(c√n): multiply top and bottom by √n. k/(a + √b): multiply top and bottom by the conjugate (a − √b).

Can you…?

I can write numbers using index notation and use the three index laws with numbers and letters.
I know that a⁰ = 1 and that a negative index means “one over”.
I can work out roots, place a root between two whole numbers, and use a calculator for powers and roots.
I can write any large or small number in standard form and convert back.
I can compare, order and calculate with numbers in standard form, with and without a calculator.
I can give exact answers for circle questions in terms of π.
(Higher) I can work with fractional indices and solve index equations.
(Higher) I can simplify, add, multiply and expand surds.
(Higher) I can rationalise a denominator, including using the conjugate.

Key terms

Base: The number that is multiplied in a power. In 3⁴ the base is 3.
Index (power, exponent): The small raised number that says how many lots of the base are multiplied. Plural: indices.
Reciprocal: One divided by a number. The reciprocal of 5 is 1/5 and a⁻¹ = 1/a.
Root: The number that gives the original when raised to a power. √25 = 5 and ³√8 = 2.
Perfect square / cube: The square or cube of a whole number: 1, 4, 9, 16… and 1, 8, 27, 64…
Standard form: A number written as A × 10ⁿ with 1 ≤ A < 10 and n a whole number.
Significant figures: The digits of a number counted from the first non-zero digit.
Exact answer: An answer that has not been rounded, such as 36π or 2√3.
Surd: A square root that is not a whole number or fraction, such as √2. It is irrational.
Irrational number: A number that cannot be written as a fraction. Its decimal never ends or repeats.
Like surds: Surds with the same number under the root, for example 3√2 and 5√2.
Rationalise the denominator: Rewrite a fraction so that its bottom is a whole number, with no surd.
Conjugate: For a + √b, the conjugate is a − √b (the sign in the middle changed). Their product is a² − b.

Total 40 marks. Allow about 40 minutes.

← Chapter 5: Percentages Chapter 7: Algebra basics →

GCSE Maths Standard Form, Indices and Surds (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Index notation and the laws of indices

Key ideas

Watch out

Exam tip

Exercise 6A Powers and the laws of indices with numbers

Exercise 6B The laws of indices with letters

Zero and negative indices

Key ideas

Watch out

Exercise 6C Zero and negative indices

Exercise 6D Index equations and mixed practice

Roots and fractional indices

Key ideas

Watch out

Higher tier: fractional indices

Key ideas (Higher)

Watch out

Exercise 6E Roots without a calculator

Exercise 6F Powers and roots with a calculator

Exercise 6G Fractional indices

Exercise 6H Fractional indices in algebra and equations

Standard form: large and small numbers

Key ideas

Watch out

Exam tip

Exercise 6I Writing numbers in standard form

Exercise 6J Standard form in context

Comparing and calculating with standard form

Key ideas

Watch out

Exam tip

Exercise 6K Comparing and ordering standard form

Exercise 6L Calculating with standard form (no calculator)

Exercise 6M Calculator questions and context

Exact answers with π

Key ideas

Watch out

Exercise 6N Exact answers with π

Surds: simplifying and adding

Key ideas

Watch out

Exercise 6O Simplifying and adding surds

Multiplying, dividing and expanding with surds

Key ideas

Watch out

Exercise 6P Multiplying, dividing and expanding with surds

Rationalising the denominator and exact answers

Key ideas

Watch out

Exercise 6Q Rationalising the denominator

Exercise 6R Exact answers with surds

Chapter summary

Can you…?

Key terms

Exercise 6Z Check-up (non-calculator)

Exercise 6X Exam-style questions