2.2 Surds and rationalising the denominator — Practise

Work through these on your own. Give every answer in its simplest exact form, not as a decimal. When a denominator contains a surd, rationalise it.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Simplify $\sqrt{50}$ .
Simplify $\sqrt{72} - \sqrt{32}$ .
Simplify $\sqrt{12} + \sqrt{75}$ .
Work out $\sqrt{8} \times \sqrt{18}$ .
Expand and simplify $\left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right)$ .
Write $\dfrac{6}{\sqrt{3}}$ with a rational denominator.
Write $\dfrac{1}{3 + \sqrt{2}}$ with a rational denominator.
Write $\dfrac{5}{\sqrt{7} - 2}$ with a rational denominator.
Expand and simplify $\left(\sqrt{5} + 2\right)^2$ .
Explain why $\sqrt{2} + \sqrt{3} > \sqrt{5}$ .
Write $\dfrac{2}{3\sqrt{2}}$ with a rational denominator.
Write $\dfrac{2 + \sqrt{3}}{1 - \sqrt{3}}$ with a rational denominator.

Answers and concise solutions

1. Simplified

Answer: Look for the square factor: $50 = 25 \times 2$ , so $\sqrt{50} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$ .

2. Simplified

Answer: Simplify each surd first: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ and $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$ . Then subtract like terms: $6\sqrt{2} - 4\sqrt{2} = 2\sqrt{2}$ .

3. Simplified

Answer: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ and $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ . The surds are now like terms, so $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ .

4. The value

Answer: Multiply inside the roots: $\sqrt{8} \times \sqrt{18} = \sqrt{144} = 12$ .

5. Simplified

Answer: This is a difference of two squares: $\left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right) = 2^2 - \left(\sqrt{3}\right)^2 = 4 - 3 = 1$ . A conjugate pair always removes the surd.

6. Rationalised

Answer: Multiply the top and the bottom by $\sqrt{3}$ : $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$ .

7. Rationalised

Answer: Multiply the top and the bottom by the conjugate $3 - \sqrt{2}$ . The denominator becomes $9 - 2 = 7$ , so $\dfrac{1}{3 + \sqrt{2}} = \dfrac{3 - \sqrt{2}}{7}$ .

8. Rationalised

Answer: Multiply the top and the bottom by the conjugate $\sqrt{7} + 2$ . The denominator becomes $7 - 4 = 3$ , so $\dfrac{5}{\sqrt{7} - 2} = \dfrac{5\left(\sqrt{7} + 2\right)}{3}$ .

9. Simplified

Answer: Expand the bracket: $\left(\sqrt{5} + 2\right)^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$ . The middle term is $2 \times \sqrt{5} \times 2$ .

10. A valid reason

Answer: $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$ , so the left side is about $3.146$ , while $\sqrt{5} \approx 2.236$ . The left side is larger. Use the decimal values only to compare sizes, never as the answer itself.

11. Rationalised

Answer: Multiply the top and the bottom by $\sqrt{2}$ : $\dfrac{2}{3\sqrt{2}} = \dfrac{2\sqrt{2}}{3 \times 2} = \dfrac{\sqrt{2}}{3}$ .

12. Rationalised

Answer: Multiply the top and the bottom by the conjugate $1 + \sqrt{3}$ . The numerator is $2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}$ , and the denominator is $1 - 3 = -2$ . So the fraction is $-\dfrac{5 + 3\sqrt{3}}{2}$ .