Question paper, Set 3

$\sqrt[5]{243} = \sqrt[5]{3 \times 3 \times 3 \times 3 \times 3} = 3$

It is a rational number, so it is not a surd.

Answer: No. $\sqrt[5]{243} = 3$, a rational number.

$1^2 = 1$ and $2^2 = 4$, so $3$ is not a perfect square.

So $\sqrt{3}$ is not rational: it is a surd.

$\pi = 3.14\ldots$ is a non-terminating, non-recurring decimal: it is irrational.

But $\pi$ is not a root of a rational number, so it is not a surd.

Answer: $\sqrt{3}$ is a surd; $\pi$ is irrational but not a surd.

$5^2 = 25$ and $6^2 = 36$, so $29$ is not a perfect square.

So $\sqrt{29}$ is a surd, and $8\sqrt{29}$ is in the form $a \cdot \sqrt{b}$: a mixed surd.

Answer: Yes. It is $8$ times the surd $\sqrt{29}$.

(i) $\sqrt{11} \times \sqrt{15} = \sqrt{11 \times 15} = \sqrt{165}$

(ii) $\sqrt{37} \times \sqrt{37} = (\sqrt{37})^2 = 37$

Answer: (i) $\sqrt{165}$ (ii) $37$

(i) $\displaystyle \frac{\sqrt{28}}{\sqrt{7}} = \sqrt{\frac{28}{7}}$

$= \sqrt{4} = \sqrt{2 \times 2} = 2$

(ii) $\displaystyle \frac{\sqrt{207}}{\sqrt{23}} = \sqrt{\frac{207}{23}}$

$= \sqrt{9} = \sqrt{3 \times 3} = 3$

Answer: (i) $2$ (ii) $3$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$

So $\displaystyle \sqrt[6]{22} = 22^{\frac{1}{6}}$

Answer: $\displaystyle 22^{\frac{1}{6}}$

$(9\sqrt{2})^2 = 9^2 \times (\sqrt{2})^2$

$= 81 \times 2 = 162$

Answer: $162$

(i) $(5 - \sqrt{10})(3 - \sqrt{10}) = 15 - 5\sqrt{10} - 3\sqrt{10} + 10$

$= 25 - 8\sqrt{10}$

(ii) $(1 + \sqrt{6})(9 + \sqrt{6}) = 9 + \sqrt{6} + 9\sqrt{6} + 6$

$= 15 + 10\sqrt{6}$

Answer: (i) $25 - 8\sqrt{10}$ (ii) $15 + 10\sqrt{6}$

$(\sqrt{17} - \sqrt{5})(\sqrt{17} + \sqrt{5}) = (\sqrt{17})^2 - (\sqrt{5})^2$

$= 17 - 5 = 12$, a rational number.

So the rationalising factor is $\sqrt{17} + \sqrt{5}$ and the product is $12$.

Answer: Rationalising factor $\sqrt{17} + \sqrt{5}$; product $12$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{15}$, $y = \sqrt[3]{3}$.

Rationalising factor: $(\sqrt[3]{15})^2 + \sqrt[3]{15} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$

Check: the product is $12$, a rational number.

Answer: $(\sqrt[3]{15})^2 + \sqrt[3]{15} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$

$\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}$

$\sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}$

$3\sqrt{96} = 3 \times 4\sqrt{6} = 12\sqrt{6}$

$\sqrt{150} + \sqrt{54} + 3\sqrt{96} = 5\sqrt{6} + 3\sqrt{6} + 12\sqrt{6} = 20\sqrt{6}$

So $k = 20$

Answer: $k = 20$

$(\sqrt{19})^3 = (\sqrt{19})^2 \times \sqrt{19} = 19\sqrt{19}$

Answer: $19\sqrt{19}$

$(5\sqrt{7} - \sqrt{5})^2 = (5\sqrt{7})^2 + (\sqrt{5})^2 - 2(5\sqrt{7})(\sqrt{5})$

$(5\sqrt{7})^2 = 25 \times 7 = 175$

$2(5\sqrt{7})(\sqrt{5}) = 10\sqrt{35}$

$= 175 + 5 - 10\sqrt{35} = 180 - 10\sqrt{35}$

Answer: $180 - 10\sqrt{35}$

$(a + b)(a - b) = a^2 - b^2$

$= 20^2 - (\sqrt{14})^2 = 400 - 14 = 386$

Answer: $386$

(i) $\displaystyle \frac{14}{4 - \sqrt{2}} = \frac{14(4 + \sqrt{2})}{(4 - \sqrt{2})(4 + \sqrt{2})}$

$(4 - \sqrt{2})(4 + \sqrt{2}) = 16 - 2 = 14$

$\displaystyle = \frac{14(4 + \sqrt{2})}{14}$

$= 4 + \sqrt{2}$

(ii) $\displaystyle \frac{9}{\sqrt{14} + \sqrt{5}} = \frac{9(\sqrt{14} - \sqrt{5})}{(\sqrt{14} + \sqrt{5})(\sqrt{14} - \sqrt{5})}$

$(\sqrt{14} + \sqrt{5})(\sqrt{14} - \sqrt{5}) = 14 - 5 = 9$

$\displaystyle = \frac{9(\sqrt{14} - \sqrt{5})}{9}$

$= \sqrt{14} - \sqrt{5}$

Answer: (i) $4 + \sqrt{2}$ (ii) $\sqrt{14} - \sqrt{5}$

$\displaystyle \frac{7 + 3\sqrt{5}}{3 - \sqrt{5}} = \frac{(7 + 3\sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}$

Numerator: $21 + 7\sqrt{5} + 9\sqrt{5} + 15 = 36 + 16\sqrt{5}$

Denominator: $3^2 - (\sqrt{5})^2 = 9 - 5 = 4$

$\displaystyle \frac{36 + 16\sqrt{5}}{4} = 9 + 4\sqrt{5}$

Comparing with $a + b\sqrt{5}$: $a = 9$, $b = 4$

Answer: $a = 9$, $b = 4$

$\displaystyle = \frac{(7 + \sqrt{6})^2}{(7 - \sqrt{6})(7 + \sqrt{6})} + \frac{(7 - \sqrt{6})^2}{(7 + \sqrt{6})(7 - \sqrt{6})}$

$(7 + \sqrt{6})^2 = 49 + 6 + 14\sqrt{6} = 55 + 14\sqrt{6}$, $(7 - \sqrt{6})^2 = 55 - 14\sqrt{6}$, $(7 + \sqrt{6})(7 - \sqrt{6}) = 43$

$\displaystyle = \frac{55 + 14\sqrt{6}}{43} + \frac{55 - 14\sqrt{6}}{43}$

$\displaystyle = \frac{55 + 14\sqrt{6} + 55 - 14\sqrt{6}}{43} = \frac{110}{43}$

Answer: $\displaystyle \frac{110}{43}$

$\displaystyle \frac{9}{\sqrt{11} + \sqrt{2}} = \frac{9(\sqrt{11} - \sqrt{2})}{(\sqrt{11} + \sqrt{2})(\sqrt{11} - \sqrt{2})} = \frac{9(\sqrt{11} - \sqrt{2})}{9} = \sqrt{11} - \sqrt{2}$

$\displaystyle \frac{12}{\sqrt{14} + \sqrt{2}} = \frac{12(\sqrt{14} - \sqrt{2})}{(\sqrt{14} + \sqrt{2})(\sqrt{14} - \sqrt{2})} = \frac{12(\sqrt{14} - \sqrt{2})}{12} = \sqrt{14} - \sqrt{2}$

$\displaystyle \frac{3}{\sqrt{14} + \sqrt{11}} = \frac{3(\sqrt{14} - \sqrt{11})}{(\sqrt{14} + \sqrt{11})(\sqrt{14} - \sqrt{11})} = \frac{3(\sqrt{14} - \sqrt{11})}{3} = \sqrt{14} - \sqrt{11}$

Sum $= (\sqrt{11} - \sqrt{2}) - (\sqrt{14} - \sqrt{2}) + (\sqrt{14} - \sqrt{11})$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{4 + \sqrt{15}} = \frac{4 - \sqrt{15}}{(4 + \sqrt{15})(4 - \sqrt{15})}$

$\displaystyle = \frac{4 - \sqrt{15}}{16 - 15} = 4 - \sqrt{15}$

$\displaystyle \frac{1}{x} = 4 + \sqrt{15}$

$\displaystyle x + \frac{1}{x} = (4 - \sqrt{15}) + (4 + \sqrt{15}) = 8$

Answer: $8$

(i) $\sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}$

$\sqrt{17 + \sqrt{168}} = \sqrt{14 + 3 + 2\sqrt{14 \times 3}}$

$= \sqrt{(\sqrt{14})^2 + (\sqrt{3})^2 + 2(\sqrt{14})(\sqrt{3})}$

$= \sqrt{(\sqrt{14} + \sqrt{3})^2} = \sqrt{14} + \sqrt{3}$

(ii) $4\sqrt{15} = 2 \times 2\sqrt{15} = 2\sqrt{60}$

$\sqrt{19 - 4\sqrt{15}} = \sqrt{15 + 4 - 2\sqrt{15 \times 4}}$

$= \sqrt{(\sqrt{15})^2 + (2)^2 - 2(\sqrt{15})(2)}$

$= \sqrt{(\sqrt{15} - 2)^2} = \sqrt{15} - 2$

($\sqrt{15} > 2$, so (ii) is positive.)

Answer: (i) $\sqrt{14} + \sqrt{3}$ (ii) $\sqrt{15} - 2$