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Square roots and cube roots

Numbers · level 5, lesson 6 of 7

A question to start

A square garden has an area of 49 square metres. How long is each side? You probably said 7 m at once, because \(7 \times 7 = 49\).

Now a square garden has an area of 40 square metres. How long is each side? No whole number works: \(6 \times 6 = 36\) is too small and \(7 \times 7 = 49\) is too big. The side is somewhere between 6 m and 7 m. Can you find it to the nearest centimetre, with only multiplying and a bit of patience? You can, and this lesson shows how.

The idea

Squaring and its undo

Squaring a number means multiplying it by itself: \(7^2 = 49\). The square root goes the other way. The square root of 49 is the number that, squared, gives 49. We write it with the sign \(\sqrt{\ }\), called a radical sign:

\[\sqrt{49} = 7 \quad \text{because} \quad 7^2 = 49\]

The square root is the side of a square whose area is the number. That is where the name comes from: the side is the "root" the square grows from.

36 sides 6, √40, 7 49 40 67√40 ≈ 6.32

Numbers like \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144\) are the squares of whole numbers. They are called perfect squares, and their square roots are whole numbers. Knowing the perfect squares up to \(15^2 = 225\) by heart makes all the estimating in this lesson quick.

Positive and negative

Both \(7^2 = 49\) and \((-7)^2 = 49\) (negative times negative is positive, lesson 1). So 49 has two numbers that square to it. The sign \(\sqrt{49}\) means only the non-negative one, 7. When we want both, we write \(\pm 7\) ("plus or minus 7"). So:

  • the equation \(x^2 = 49\) has two solutions, \(x = 7\) and \(x = -7\);
  • but \(\sqrt{49}\) is just \(7\).

What about \(\sqrt{-9}\)? We need a number that squares to \(-9\). A positive number squared is positive, and a negative number squared is also positive, and \(0^2 = 0\). So no number we know squares to a negative. For now, square roots of negative numbers have no answer. (In level 7 a whole new kind of number is invented to give them one.)

Roots of fractions and decimals

Squaring a fraction squares the top and bottom: \(\left(\tfrac34\right)^2 = \tfrac9{16}\). So the square root works on top and bottom: \(\sqrt{\tfrac{9}{16}} = \tfrac34\).

For decimals, think about place value: \(0.5^2 = 0.25\), so \(\sqrt{0.25} = 0.5\). And \(1.2^2 = 1.44\), so \(\sqrt{1.44} = 1.2\). A useful check: the square root of a number between 0 and 1 is bigger than the number, because squaring a number between 0 and 1 makes it smaller.

Estimating a square root: trap it, then squeeze

To estimate \(\sqrt{40}\):

Step 1. Trap it between perfect squares. \(36 < 40 < 49\), so \(\sqrt{36} < \sqrt{40} < \sqrt{49}\), that is, \(6 < \sqrt{40} < 7\). (A bigger square has a bigger side, so the order of the areas is the order of the sides.)

Step 2. Guess where in the gap. 40 is 4 of the way from 36 towards 49, out of 13. So a first guess is about \(6 + \tfrac{4}{13} \approx 6.3\). (This guess is a little low, because square roots grow more slowly as numbers get bigger, but it is a good start.)

Step 3. Squeeze by guess and check. Square your guesses:

guess square too big or too small?
6.3 39.69 too small
6.4 40.96 too big
6.32 39.9424 too small
6.33 40.0689 too big

So \(6.32 < \sqrt{40} < 6.33\). And 40 is closer to 39.9424 than to 40.0689, so \(\sqrt{40} \approx 6.32\). A calculator gives \(6.32455\ldots\) Every new row gives one more correct decimal place. In level 6 you will meet a much faster method, nearly 4000 years old.

Will the decimals of \(\sqrt{40}\) ever stop or repeat? They will not. Numbers like this are the subject of level 6.

A useful rule: roots of products

\(\sqrt{4 \times 9} = \sqrt{36} = 6\), and \(\sqrt4 \times \sqrt9 = 2 \times 3 = 6\). The same! In general, for numbers that are not negative,

\[\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\]

Why? \(\sqrt a \times \sqrt b\) is a number whose square is \((\sqrt a \sqrt b)^2 = (\sqrt a)^2 (\sqrt b)^2 = a \times b\) (Law 4 from Powers). And it is not negative. So it is the square root of \(ab\).

This lets you find roots of big numbers in your head: \(\sqrt{900} = \sqrt{9} \times \sqrt{100} = 3 \times 10 = 30\). In level 6 you will use this rule to simplify roots such as \(\sqrt{72}\).

But there is no such rule for adding: \(\sqrt{9 + 16} = \sqrt{25} = 5\), while \(\sqrt9 + \sqrt{16} = 3 + 4 = 7\).

Cube roots

The cube root of a number is the number that, cubed, gives it. It is the edge of a cube with that volume. We write it \(\sqrt[3]{\ }\):

\[\sqrt[3]{27} = 3 \quad \text{because} \quad 3^3 = 27\]

The perfect cubes are \(1, 8, 27, 64, 125, 216, 343, 512, 729, 1000\).

Cube roots have one big difference from square roots: negative numbers have cube roots. \((-2)^3 = -8\), so \(\sqrt[3]{-8} = -2\). An odd number of negative factors gives a negative answer, so every number, positive or negative, has exactly one cube root.

Cube roots can be estimated the same way. \(64 < 100 < 125\), so \(4 < \sqrt[3]{100} < 5\). Then \(4.6^3 = 97.336\) and \(4.7^3 = 103.823\), so \(\sqrt[3]{100}\) is between 4.6 and 4.7 (it is \(4.64\ldots\)).

Worked examples

1. Exact roots. Find \(\sqrt{144}\), \(\sqrt{0.36}\) and \(\sqrt{\tfrac{25}{49}}\).

  1. \(12^2 = 144\), so \(\sqrt{144} = 12\).
  2. \(0.6^2 = 0.36\), so \(\sqrt{0.36} = 0.6\).
  3. \(\sqrt{\tfrac{25}{49}} = \tfrac{\sqrt{25}}{\sqrt{49}} = \tfrac57\).

2. Estimate \(\sqrt{20}\) to two decimal places.

  1. Trap: \(16 < 20 < 25\), so \(4 < \sqrt{20} < 5\), and 20 is a bit under halfway.
  2. Squeeze: \(4.4^2 = 19.36\) (small), \(4.5^2 = 20.25\) (big). So it is between 4.4 and 4.5.
  3. Squeeze more: \(4.47^2 = 19.9809\) (small), \(4.48^2 = 20.0704\) (big). 20 is much closer to 19.9809.
  4. So \(\sqrt{20} \approx 4.47\).

3. Fencing a square plot. A square plot has area 50 m². How much fence goes round it?

  1. Side \(= \sqrt{50}\). Since \(7^2 = 49\), the side is just over 7 m. In fact \(7.07^2 = 49.9849\) and \(7.08^2 = 50.1264\), so the side is about 7.07 m.
  2. Fence \(= 4 \times 7.07 \approx\) 28.3 m. (Buy a little more, to be safe.)

4. Cube roots. Find \(\sqrt[3]{-125}\) and \(\sqrt[3]{0.001}\).

  1. \((-5)^3 = -125\), so \(\sqrt[3]{-125} = -5\).
  2. \(0.1^3 = 0.001\), so \(\sqrt[3]{0.001} = 0.1\).

5. Use the product rule. Find \(\sqrt{1600}\) and \(\sqrt{0.09}\).

  1. \(\sqrt{1600} = \sqrt{16} \times \sqrt{100} = 4 \times 10 = 40\).
  2. \(\sqrt{0.09} = 0.3\), because \(0.3 \times 0.3 = 0.09\). (Or: \(0.09 = \tfrac{9}{100}\), so the root is \(\tfrac{3}{10}\).)

Try it

  1. Find: (a) \(\sqrt{81}\) (b) \(\sqrt{169}\) (c) \(\sqrt{0.49}\) (d) \(\sqrt{\tfrac{4}{81}}\)
  2. Between which two whole numbers is \(\sqrt{75}\)? Which one is it closer to?
  3. Estimate \(\sqrt{30}\) to one decimal place by guess and check.
  4. A square tile has area 2.25 m². How long is each side?
  5. Find: (a) \(\sqrt[3]{64}\) (b) \(\sqrt[3]{-27}\) (c) \(\sqrt[3]{1\,000\,000}\)
  6. Solve \(x^2 = 121\). Then say what \(\sqrt{121}\) is, and why the answers are different.
  7. Which is bigger, \(\sqrt{50}\) or \(\sqrt[3]{300}\)? Explain without a calculator.
  8. Explain why \(\sqrt{-16}\) has no answer among the numbers we know, but \(\sqrt[3]{-64}\) does.
Answers
  1. (a) 9. (b) 13. (c) 0.7, since \(0.7^2 = 0.49\). (d) \(\tfrac29\).
  2. \(64 < 75 < 81\), so between 8 and 9. 75 is 6 away from 81 and 11 away from 64, so it is closer to 9. (It is \(8.66\ldots\))
  3. \(25 < 30 < 36\), so between 5 and 6. \(5.4^2 = 29.16\) and \(5.5^2 = 30.25\), so between 5.4 and 5.5. 30 is closer to 30.25 (0.25 away) than to 29.16 (0.84 away), so about 5.5. (It is \(5.477\ldots\), which also rounds to 5.5.)
  4. \(1.5^2 = 2.25\), so 1.5 m.
  5. (a) 4. (b) \(-3\). (c) 100, since \(100^3 = 1\,000\,000\).
  6. \(x = 11\) or \(x = -11\), because both square to 121. But \(\sqrt{121} = 11\) only: the root sign means the non-negative root.
  7. \(\sqrt{50} > \sqrt{49} = 7\). And \(\sqrt[3]{300} < \sqrt[3]{343} = 7\). So \(\sqrt{50}\) is bigger.
  8. Any number squared is positive or zero (positive times positive and negative times negative are both positive), so nothing squares to \(-16\). But \((-4)^3 = -64\), because three negative factors give a negative, so \(\sqrt[3]{-64} = -4\).

Common mistake

Halving instead of rooting. \(\sqrt{16}\) is 4, not 8. The square root asks "what times itself gives 16?", not "what times 2?". Check by squaring: \(8^2 = 64\), not 16.

Splitting a root over a sum. \(\sqrt{9 + 16}\) is \(\sqrt{25} = 5\), not \(3 + 4 = 7\). The rule \(\sqrt{ab} = \sqrt a \sqrt b\) is for products only. Always work out what is under the root sign first.

Where it leads