Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Number
Practise Indices and roots. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Squares, cubes and roots, powers of numbers, and the laws of indices, including zero and negative indices and, on the Higher tier, fractional indices. Expect short non-calculator questions such as 'write down the value of' or 'write as a single power', worth 1 to 3 marks.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
1
Recall square numbers and square rootsKnow the squares up to \(15^2 = 225\) and their square roots, e.g. \(\sqrt{144} = 12\).
2
Recall cube numbers and cube rootsKnow 1, 8, 27, 64, 125 and 1000, e.g. \(\sqrt[3]{64} = 4\).
3
Recognise powers of 2, 3, 4 and 5E.g. \(2^5 = 32\), \(3^4 = 81\), \(4^3 = 64\) and \(5^3 = 125\).
4
Use the laws of indices with numbersAdd indices to multiply, subtract to divide, multiply for a power of a power: \((3^4)^2 = 3^8\).
5
Use zero and negative indices\(a^0 = 1\) and \(a^{-n} = \frac{1}{a^n}\), e.g. \(2^{-3} = \frac{1}{8}\).
Learn the powers of 2 up to \(2^{10} = 1024\): they turn up in many index questions.
Where marks are lost
Multiplying the bases: \(3^4 \times 3^2 = 3^6\), not \(9^6\).
Multiplying the indices instead of adding: \(5^3 \times 5^4 = 5^7\), not \(5^{12}\).
Thinking a negative index makes the number negative: \(2^{-3} = \frac{1}{8}\), not −8.
Writing \(7^0 = 0\) or \(7^0 = 7\). It is 1.
Working out \(2 \times 3^2\) as \(6^2 = 36\). The power comes first: 2 × 9 = 18.
Exam technique
'Write down' means you can give the answer without working, but it must be exact.
If asked to write as a power (e.g. 'as a power of 7'), leave the answer as \(7^8\); don't work out its value.
Rewrite every number as a power of the same prime before using the laws, e.g. \(8 = 2^3\) and \(27 = 3^3\).
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
(a) Write down the value of \(6^{-2}\)[1]
(b) Write down the value of \(\left(\frac{1}{9}\right)^{0}\)[1]
(c) Write down the value of \(49^{\frac{1}{2}}\)[1]
Show the answer and mark scheme
(a)Answer: \(\frac{1}{36}\)
B1 for \(\frac{1}{36}\) oe
Worked solution: \(6^{-2} = \frac{1}{6^2} = \frac{1}{36}\)
(b)Answer: \(1\)
B1 for 1
Worked solution: Any non-zero number to the power 0 is 1.
(c)Answer: \(7\)
B1 for 7
Worked solution: \(49^{\frac{1}{2}} = \sqrt{49} = 7\)
Question 2Medium3 marks
Mia works out the value of \(8^{-\frac{2}{3}}\). Here is her working. \(8^{-\frac{2}{3}} = -\left(8^{\frac{2}{3}}\right) = -\left(\sqrt[3]{8}\right)^2 = -4\)
(a) Explain Mia's mistake.[1]
(b) Work out the correct value of \(8^{-\frac{2}{3}}\).[2]
Show the answer and mark scheme
(a)Answer: A negative power means the reciprocal, not a negative number: \(8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}}\).
C1 for explaining that the negative index means 'one over' (the reciprocal), not a negative answer
Worked solution: \(x^{-n} = \frac{1}{x^n}\), so the minus sign in the index gives a reciprocal. Mia made the answer negative instead.
(b)Answer: \(\frac{1}{4}\)
M1 for \(\frac{1}{8^{\frac{2}{3}}}\) or \(8^{\frac{2}{3}} = 4\) or \(\left(\frac{1}{2}\right)^2\)
A1 for \(\frac{1}{4}\) oe
Worked solution: \(8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\), so \(8^{-\frac{2}{3}} = \frac{1}{4}\).