Calculations with Square Roots
Calculations with Square Roots gets you ready for Pythagoras' theorem in NSW Year 8 maths (NESA outcome MA4‑IND‑C‑01). A root sign works like a bracket, so in \(\sqrt{5^2 + 12^2}\) you square, add and only then take the root, which gives \(13\) and not \(5 + 12\).
You will evaluate roots of large numbers and decimals such as \(\sqrt{2.25}\), find cube roots, round answers to one decimal place and decide whether a root is exact or a surd. You will also trap a root between two whole numbers and find a missing number under a root sign. A square root gives the side of a square photo from its area. It is part of Year 8 Pythagoras.
Theory
The square root sign works like a bracket: work out everything under it first, then take the root. \(\sqrt{4^2 + 3^2} = \sqrt{25} = 5\), not \(4 + 3\). This Year 8 maths page shows how to find large and decimal roots, cube roots, exact answers and surds, and how to estimate a root. These skills are used with Pythagoras’ theorem.
Under the root first. In \(\sqrt{a^2 + b^2}\), square each number, add, and only then take the root. The answer is not \(a + b\).
Big perfect squares. Trap the number between two multiples of \(10\), then use its last digit. Or split it into square factors, because \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\): \(\sqrt{4900} = \sqrt{49} \times \sqrt{100} = 7 \times 10 = 70\).
Decimals. Count the decimal places: \(0.8 \times 0.8 = 0.64\), so \(\sqrt{0.64} = 0.8\). Two decimal places under the root give one in the answer.
Cube roots. Write the number as a product of prime factors, then split them into three equal groups. Each group is the cube root.
Exact or approximate. If the root works out exactly (a whole number, a fraction or a decimal that stops) it is rational. If the decimal never stops, the root is a surd, and the exact answer keeps the root sign. A rounded decimal is only an approximation.
Estimating. Trap the number between the square numbers either side, then take roots: \(49 < 55 < 64\), so \(7 < \sqrt{55} < 8\). For a cube root, use cube numbers: \(64 < 100 < 125\), so \(\sqrt[3]{100}\) is between \(4\) and \(5\). Going backwards: squaring undoes a square root; if \(\sqrt{n} = 9\), then \(n = 81\). With a missing number inside, square both sides, subtract, then take the square root. Order of operations: roots first (like powers), then \(\times\) and \(\div\), then \(+\) and \(-\).
Exact answers and rounded answers:
| Root | Exact answer | To 1 decimal place |
|---|---|---|
| \(\sqrt{1600}\) | \(40\) | \(40.0\) |
| \(\sqrt{0.04}\) | \(0.2\) | \(0.2\) |
| \(\sqrt{30}\) | \(\sqrt{30}\), a surd | \(5.5\) |
| \(\sqrt[3]{8000}\) | \(20\) | \(20.0\) |
| \(\sqrt[3]{20}\) | \(\sqrt[3]{20}\), a surd | \(2.7\) |
How to evaluate a root of a sum or difference
- Square each number under the root.
- Add or subtract under the root.
- Take the square root: exact if it is a square number, otherwise keep the surd or round.
How to estimate a square root
- Find the square numbers just below and just above.
- Take their roots: the answer lies between them.
(i) \(\sqrt{4^2 + 3^2}\)
(ii) \(\sqrt{25^2 - 24^2}\)
| \(\sqrt{4^2 + 3^2}\) | \(=\) | \(\sqrt{16 + 9} = \sqrt{25}\) |
| \(\) | \(=\) | \(5\) |
| \(\sqrt{25^2 - 24^2}\) | \(=\) | \(\sqrt{625 - 576} = \sqrt{49}\) |
| \(\) | \(=\) | \(7\) |
(i) \(5\)
(ii) \(7\)
(i) \(\sqrt{5^2 + 4^2}\)
(ii) \(\sqrt{5^2 - 2^2}\)
| \(\sqrt{5^2 + 4^2}\) | \(=\) | \(\sqrt{25 + 16} = \sqrt{41}\) |
| \(\sqrt{5^2 - 2^2}\) | \(=\) | \(\sqrt{25 - 4} = \sqrt{21}\) |
(i) \(\sqrt{41}\) (a surd) \(\approx 6.4\)
(ii) \(\sqrt{21}\) (a surd) \(\approx 4.6\)
(i) \(\sqrt{1444}\)
(ii) \(\sqrt{0.81}\)
(i) \(30^2 = 900 < 1444 < 1600 = 40^2\). \(1444\) ends in \(4\), so the root ends in \(2\) or \(8\); and \(35^2 = 1225 < 1444\), so try \(38\).
(ii) \(9 \times 9 = 81\); two decimal places under the root give one in the answer.
| \(38 \times 38\) | \(=\) | \(1444\) |
| \(0.9 \times 0.9\) | \(=\) | \(0.81\) |
(i) \(38\)
(ii) \(0.9\)
(i) \(\sqrt{20^2 + 21^2}\)
(ii) \(\sqrt{29^2 - 21^2}\)
| \(20^2 + 21^2\) | \(=\) | \(400 + 441 = 841\) |
| \(29^2 - 21^2\) | \(=\) | \(841 - 441 = 400\) |
(i) \(\sqrt{841} = 29\)
(ii) \(\sqrt{400} = 20\). Note \(29 - 21 = 8\) is not the answer.
(i) exactly
(ii) between two whole numbers
(iii) to \(1\) decimal place
| \(6^2 + 7^2\) | \(=\) | \(36 + 49 = 85\) |
(i) \(\sqrt{85}\), a surd
(ii) \(9^2 = 81 < 85 < 100 = 10^2\), so between \(9\) and \(10\)
(iii) \(\sqrt{85} = 9.2195\ldots \approx 9.2\)
\(2744 = 2^3 \times 7^3\), so \(\sqrt[3]{2744} = 2 \times 7 = 14\). \(11^2 = 121\), so \(\sqrt{1.21} = 1.1\).
| \(14 - 4 \times 1.1\) | \(=\) | \(14 - 4.4\) |
| \(\) | \(=\) | \(9.6\) |
\(9.6\)
(ii) What is the smallest whole number \(n\) for which \(\sqrt{n}\) is greater than \(8.5\)?
(iii) Find the whole number: \(\sqrt{\square^2 + 5^2} = 13\).
| \(8.5 \times 8.5\) | \(=\) | \(72.25\) |
(i) \(\sqrt{300} = 17.32\ldots \approx 17.3\) m
(ii) \(\sqrt{n} > 8.5\) means \(n > 72.25\), so \(n = 73\).
(iii) Square both sides: \(\square^2 + 25 = 169\), so \(\square^2 = 144\) and \(\square = \sqrt{144} = 12\).
Common pitfalls
Frequently asked questions
How do you evaluate the square root of a sum?
Work out the sum first, then take the root: \(\sqrt{4^2 + 3^2} = \sqrt{25} = 5\).
What is a surd?
A root that is not a whole number, fraction or terminating decimal, such as \(\sqrt{30}\). Its decimal never stops.
How do you estimate a square root?
Trap the number between two square numbers. \(49 < 55 < 64\), so \(\sqrt{55}\) is between \(7\) and \(8\).
How do you find a cube root from prime factors?
Split the prime factors into three equal groups: \(2744 = (2 \times 7)^3\), so the cube root is \(14\).
How do you find the square root of a decimal?
Use the whole-number root and halve the decimal places: \(\sqrt{0.81} = 0.9\).