Powers and roots
Section: Number 1 | Syllabus: Cambridge IGCSE Mathematics (0580)
Squares and Square Roots
Squaring a number and taking a square root undo each other. Every positive number actually has two square roots, but the root symbol itself always points to just one of them.
- To square a number, multiply it by itself: n^2 = n × n
- Every positive number has two square roots - one positive and one negative
- e.g. both 9² = 81 and (-9)² = 81, so 81 has square roots 9 and -9
- The symbol √(\ ) refers only to the positive square root
- To show both square roots at once, use ±: the square roots of 81 are ±√(81) = ± 9
- Negative numbers have no real square root, since squaring any positive or negative number always gives a positive result
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 | 225 |
These squares (and their roots, read the table backwards) should be recalled without a calculator
Worked Example: Squaring and Square-Rooting Decimals
- 0.4^2 = 0.16
- 1.2^2 = 1.44
- Question: Work out √(0.0064).
- Step 1: Write the decimal as a fraction over a power of 10: 0.0064 = (64/10\,000)
- Step 2: Take the square root of the numerator and denominator separately: √(64/10\,000) = (8/100)
- Answer: √(0.0064) = 0.08
Common Mistakes
MistakeGiving only the positive root when solving an equation like x² = 49, writing x = 7 only
Fixa squared equation has two solutions: x = ±7. The √ symbol alone means "positive root", but solving x² = 49 needs both
MistakeDoubling a decimal instead of squaring it, e.g. thinking 0.4² = 0.8
Fixsquaring means multiplying by itself, not by 2: 0.4^2 = 0.4 × 0.4 = 0.16
Cubes and Cube Roots
Cubing and cube-rooting work the same way as squaring and square-rooting, but with one important difference: negative numbers behave perfectly well when cubed.
- To cube a number, multiply it by itself three times: n^3 = n × n × n
- Unlike squaring, cubing a negative number gives a negative result: (-3)^3 = -27
- This means every number, positive or negative, has exactly one real cube root
- The cube root of a negative number is itself negative: √[3]-27 = -3
| n | 1 | 2 | 3 | 4 | 5 | 10 |
|---|---|---|---|---|---|---|
| n³ | 1 | 8 | 27 | 64 | 125 | 1000 |
These cubes (and their roots, read the table backwards) should be recalled without a calculator
Worked Example: Cubes and Cube Roots
- 4^3 = 64
- √[3]125 = 5
- Question: Work out √[3]-64.
- Step 1: Find the positive number that cubes to give 64: 4^3 = 64
- Step 2: Since the number under the root is negative, the cube root is also negative: (-4)^3 = -64
- Answer: √[3]-64 = -4
Common Mistakes
MistakeAssuming a negative number has no cube root, by analogy with square roots
Fixsquare roots of negative numbers do not exist, but cube roots of negative numbers do exist, and are themselves negative: √[3]-64 = -4
MistakeWriting ± in front of a cube root, by analogy with square roots, e.g. writing √[3]125 = ± 5
Fixa cube root has only one real answer, so no ± is needed: √[3]125 = 5 only
Combining Powers and Roots
Squares, cubes and other powers and roots often appear together in the same calculation. Working through each part in the right order keeps the calculation manageable.
- Higher powers and roots follow the same pattern: a 4th power multiplies a number by itself 4 times, and a 4th root asks "what number, raised to the power 4, gives this result?"
- When a calculation mixes powers, roots and other operations, work out each power or root first, then combine the results
Worked Example: Evaluating a Mixed Expression
- Question: Work out 4^2 × √[3]27.
- Step 1: Evaluate the square: 4^2 = 16
- Step 2: Evaluate the cube root: √[3]27 = 3
- Step 3: Multiply the results: 16 × 3 = 48
- Answer: 48
Worked Example: A Fourth Root
- Question: Work out √[4]81.
- Step 1: Ask "what number, raised to the power 4, gives 81?"
- Step 2: Check 3: 3^4 = 3 × 3 × 3 × 3 = 81
- Answer: √[4]81 = 3
Common Mistakes
MistakeCombining the power and the root into one step without working each part out separately, leading to arithmetic slips
Fixevaluate each power or root on its own first, then combine the separate results with the remaining operation
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