Power and Roots
Power and Roots in NSW Year 8 maths widens the base of a power to decimals, fractions and negative numbers, and takes roots of decimals (NESA outcome MA4‑IND‑C‑01). Brackets matter: \((-3)^2 = 9\) but \(-3^2 = -9\), and \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\).
You will evaluate powers such as \(0.3^2\) and \(\left(\frac{2}{3}\right)^3\), roots like \(\sqrt{1.44}\) and \(\sqrt[3]{0.027}\), and use \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\). You will apply the order of operations, find a missing number and spot last-digit patterns. The side of a square courtyard, the edge of a cube-shaped water tank and a chain message all involve powers and roots. It is part of Year 8 Working with Numbers.
Theory
In \(a^3 = a \times a \times a\), \(a\) is the base and \(3\) is the index (or power). A square root undoes squaring: \(\sqrt{25} = \sqrt{5 \times 5} = 5\). This Year 8 maths page uses decimal, fraction and negative bases, and roots of decimals and products.
A number of the form \(a \times a \times a\) can be written \(a^3\): \(a\) is the base and \(3\) is the index. Similarly \(p^4 = p \times p \times p \times p\). Base and index cannot be swapped: \(2^5 = 32\) but \(5^2 = 25\).
Square roots. To find \(\sqrt{25}\), find the number that multiplied by itself gives \(25\): \(\sqrt{25} = \sqrt{5 \times 5} = 5\). In general \(\sqrt{a \times a} = \sqrt{a^2} = a\) for a positive number \(a\).
A decimal base. Multiply the digits, then count the places: \(1.4^2\) comes from \(14^2 = 196\) with \(2\) places, so \(1.96\). A fraction base: the index goes on the top and the bottom: \(\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}\). A negative base: \((-3)^2 = 9\) but \(-3^2 = -9\).
Roots of a decimal. Use the whole-number fact and share the places out: \(\sqrt{0.81} = 0.9\), because \(0.9 \times 0.9 = 0.81\). A square root halves the number of decimal places; a cube root divides it by \(3\).
Roots of a product. \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\), so \(\sqrt{25 \times 9} = 5 \times 3 = 15\). This does not work over \(+\) or \(-\): \(\sqrt{25} + \sqrt{9} = 8\), but \(\sqrt{25 + 9} = \sqrt{34}\).
Estimating a root. Find the square numbers either side: \(49 < 55 < 64\), so \(7 < \sqrt{55} < 8\) (about \(7.42\)). Order of operations: brackets, then powers and roots, then \(\times\) and \(\div\), then \(+\) and \(-\). Missing numbers: undo each step: \(4 \times {?}^2 + \sqrt{25} = 41\) gives \({?}^2 = 9\), so \({?} = 3\).
Last digits. The last digits of \(2^1, 2^2, 2^3, \ldots\) cycle \(2, 4, 8, 6\). Since \(20 \div 4\) leaves no remainder, \(2^{20}\) ends in \(6\). Repeated growth: a message passed on to \(2\) new people each round reaches \(2^n\) people in round \(n\).
Whole-number facts give decimal powers and roots:
| Whole fact | Decimal | Root |
|---|---|---|
| \(4^2 = 16\) | \(0.4^2 = 0.16\) | \(\sqrt{0.16} = 0.4\) |
| \(9^2 = 81\) | \(0.9^2 = 0.81\) | \(\sqrt{0.81} = 0.9\) |
| \(13^2 = 169\) | \(1.3^2 = 1.69\) | \(\sqrt{1.69} = 1.3\) |
| \(7^3 = 343\) | \(0.7^3 = 0.343\) | \(\sqrt[3]{0.343} = 0.7\) |
How to evaluate a power
- Write it as repeated multiplication (watch the brackets).
- Multiply; for decimals, count the places; for fractions, do the top and bottom.
- Decide the sign: a negative base with an odd index is negative.
How to find a root
- Look for a whole-number fact, or split into a product of squares or cubes.
- Share the decimal places: half for a square root, a third for a cube root.
- Check by multiplying back.
| \(3^2\) | \(=\) | \(3 \times 3 = 9\) |
| \(2^3\) | \(=\) | \(2 \times 2 \times 2 = 8\) |
| \(3^2 + 2^3\) | \(=\) | \(9 + 8 = 17\) |
\(17\)
(i) \(\sqrt{25 \times 9}\)
(ii) \(\sqrt{25} + \sqrt{9}\)
| \(\sqrt{25 \times 9}\) | \(=\) | \(\sqrt{5 \times 5} \times \sqrt{3 \times 3}\) |
| \(\) | \(=\) | \(5 \times 3 = 15\) |
| \(\sqrt{25} + \sqrt{9}\) | \(=\) | \(5 + 3 = 8\) |
(i) \(15\)
(ii) \(8\)
(i) \(1.4^2\)
(ii) \(\left(\dfrac{2}{5}\right)^3\)
(iii) \((-0.6)^2\)
| \(14 \times 14\) | \(=\) | \(196\) |
| \(\left(\dfrac{2}{5}\right)^3\) | \(=\) | \(\dfrac{2 \times 2 \times 2}{5 \times 5 \times 5} = \dfrac{8}{125}\) |
| \(6 \times 6\) | \(=\) | \(36\) |
(i) \(2\) places: \(1.96\)
(ii) \(\dfrac{8}{125}\)
(iii) Even index, \(2\) places: \(0.36\)
(i) \(\sqrt{0.64}\)
(ii) \(\sqrt[3]{1.331}\)
| \(0.8 \times 0.8\) | \(=\) | \(0.64\) |
| \(1.1 \times 1.1 \times 1.1\) | \(=\) | \(1.331\) |
(i) \(0.8\) (\(2\) places become \(1\))
(ii) \(1.1\) (\(3\) places become \(1\))
(i) \(4 \times 3^3 - \sqrt{100 \times 4}\)
(ii) \(\left(\sqrt{81} - \sqrt[3]{8}\right)^3\)
| \(4 \times 27 - 20\) | \(=\) | \(108 - 20 = 88\) |
| \((9 - 2)^3\) | \(=\) | \(7^3 = 343\) |
(i) \(3^3 = 27\) and \(\sqrt{100 \times 4} = 10 \times 2 = 20\), so \(88\)
(ii) \(\sqrt{81} = 9\), \(\sqrt[3]{8} = 2\), so \(343\)
(i) How long is each side?
(ii) How much braid goes around all four edges?
(iii) It folds into a cube-shaped bag of volume \(0.064\) m\(^3\). Find the edge.
| \(2.6 \times 2.6\) | \(=\) | \(6.76\) |
| \(4 \times 2.6\) | \(=\) | \(10.4\) |
| \(0.4 \times 0.4 \times 0.4\) | \(=\) | \(0.064\) |
(i) \(\sqrt{6.76} = 2.6\) m
(ii) \(10.4\) m
(iii) \(\sqrt[3]{0.064} = 0.4\) m
(ii) Find the positive number: \(4 \times {?}^2 + \sqrt{25} = 41\).
| \(4 \times {?}^2\) | \(=\) | \(41 - 5 = 36\) |
| \({?}^2\) | \(=\) | \(9\) |
(i) \(49 < 55 < 64\), so between \(7\) and \(8\)
(ii) \({?} = 3\). Check: \(4 \times 9 + 5 = 41\).
Common pitfalls
Frequently asked questions
What do the base and the index mean?
In \(5^3\) the base \(5\) is multiplied by itself \(3\) times: \(5 \times 5 \times 5 = 125\).
How do you find the square root of a decimal?
Use the whole-number fact and halve the decimal places: \(\sqrt{0.49} = 0.7\).
How do you square a fraction?
Square the top and the bottom: the square of \(\dfrac{2}{3}\) is \(\dfrac{4}{9}\).
Can you split a square root over a product?
Yes: \(\sqrt{4 \times 9} = 2 \times 3 = 6\). But not over a sum.
How do you estimate a square root?
Find the square numbers either side: \(\sqrt{30}\) is between \(5\) and \(6\).