Why \(\sqrt{50}\) is \(5\sqrt2\)
Tonight's idea
Pythagoras hands you lengths like \(\sqrt{50}\): the positive number whose square is 50. A calculator says 7.0710678…, which is rounded and hard to recognise. There is an exact form that is shorter.
A square of area 50 has side \(\sqrt{50}\). Cut it into a 5 by 5 grid. Each of the 25 small squares has area 2, so each has side \(\sqrt2\). Five of those sides make the big side: \(\sqrt{50} = 5\sqrt2\).
\(\sqrt{ab} = \sqrt a \times \sqrt b\) for positive \(a\) and \(b\).
Why? Square the right-hand side: \(\sqrt a \sqrt b \times \sqrt a \sqrt b = a \times b\). So \(\sqrt a \sqrt b\) is a positive number whose square is \(ab\), and that is exactly what \(\sqrt{ab}\) means.
To simplify a root, split off a square factor: \(\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt2 = 5\sqrt2\).
Roots split over multiplying, not over adding: \(\sqrt{9 + 16} = 5\), but \(\sqrt9 + \sqrt{16} = 7\). What you can do is collect like roots, just as \(3x + 5x = 8x\): \(3\sqrt2 + 5\sqrt2 = 8\sqrt2\).
Watch one: a right triangle with legs 4 and 8
By Pythagoras the longest side is \(\sqrt{4^2 + 8^2} = \sqrt{80}\).
The biggest square that divides 80 is 16, since \(80 = 16 \times 5\). So \(\sqrt{80} = \sqrt{16} \times \sqrt5 = 4\sqrt5\).
Spotted only the 4? Then \(\sqrt{80} = 2\sqrt{20}\), and \(20 = 4 \times 5\) gives \(2 \times 2\sqrt5 = 4\sqrt5\) again. Keep going until no square factor is left under the root.
Check: \((4\sqrt5)^2 = 16 \times 5 = 80\). The short form also shows something the decimal hides: this side is exactly 4 times \(\sqrt5\), the longest side of the triangle with legs 1 and 2.
Your turn
- Simplify: (a) \(\sqrt{12}\); (b) \(\sqrt{45}\); (c) \(\sqrt{72}\); (d) \(\sqrt{200}\).
- Simplify: (a) \(\sqrt{18} + \sqrt8\); (b) \(\sqrt{75} - \sqrt{27}\); (c) \(\sqrt6 \times \sqrt{15}\).
- (a) A square has side 7. Find its diagonal exactly, in simplest form. (b) Another square has diagonal 10. Find its side exactly.
- No calculator. (a) Which is bigger, \(3\sqrt5\) or \(2\sqrt{11}\)? Hint: run the rule backwards and put everything under the root. (b) True or false: \(\sqrt2 + \sqrt8 = \sqrt{18}\). Give a reason.
Stretch: cubes instead of squares
\(\sqrt[3]{n}\), the cube root, is the number whose cube is \(n\); so \(\sqrt[3]{27} = 3\). The same rule holds: \(\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}\). Simplify \(\sqrt[3]{54}\). Then write \(\sqrt[3]{16} + \sqrt[3]{54}\) as a single cube root.
Answers
- (a) \(\sqrt{4 \times 3} = 2\sqrt3\). (b) \(\sqrt{9 \times 5} = 3\sqrt5\). (c) \(\sqrt{36 \times 2} = 6\sqrt2\). (d) \(\sqrt{100 \times 2} = 10\sqrt2\).
- (a) \(3\sqrt2 + 2\sqrt2 = 5\sqrt2\). (b) \(5\sqrt3 - 3\sqrt3 = 2\sqrt3\). (c) \(\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}\).
- (a) \(\sqrt{7^2 + 7^2} = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt2\). (b) If the side is \(s\), then \(s^2 + s^2 = 100\), so \(s^2 = 50\) and \(s = \sqrt{50} = 5\sqrt2\), about 7.07.
- (a) \(3\sqrt5 = \sqrt9 \times \sqrt5 = \sqrt{45}\) and \(2\sqrt{11} = \sqrt{44}\). So \(3\sqrt5\) is bigger, but only just. (b) True: \(\sqrt8 = 2\sqrt2\), so the left side is \(3\sqrt2 = \sqrt{9 \times 2} = \sqrt{18}\). It is true because of like roots, not because of adding under the root: \(2 + 8\) is not 18.
Stretch. Look for cube factors: \(54 = 27 \times 2\), so \(\sqrt[3]{54} = 3\sqrt[3]{2}\). Likewise \(16 = 8 \times 2\) gives \(\sqrt[3]{16} = 2\sqrt[3]{2}\). The sum is \(5\sqrt[3]{2} = \sqrt[3]{125 \times 2} = \sqrt[3]{250}\).