Question paper, Set 8

$3^2 = 9$ and $4^2 = 16$, so $14$ is not a perfect square.

So $\sqrt{14}$ is not a rational number; it is a surd.

Answer: Yes. $14$ is not a perfect square, so $\sqrt{14}$ is not rational.

$2^3 = 8$ and $3^3 = 27$, so $26$ is not a perfect cube.

So $\sqrt[3]{26}$ is not rational: it is a surd.

$e = 2.71\ldots$ is a non-terminating, non-recurring decimal: it is irrational.

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

Answer: $\sqrt[3]{26}$ is a surd; $e$ is irrational but not a surd.

$\sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = 10$

$4\sqrt[3]{1000} = 4 \times 10 = 40$, which is rational, so it is not a surd.

Answer: No. $4\sqrt[3]{1000} = 40$, a rational number.

$\sqrt{12} \times \sqrt{75} = \sqrt{12 \times 75}$

$= \sqrt{900} = \sqrt{30 \times 30}$

$= 30$

Answer: $30$

(i) $\displaystyle \frac{\sqrt{360}}{\sqrt{10}} = \sqrt{\frac{360}{10}}$

$= \sqrt{36} = \sqrt{6 \times 6} = 6$

(ii) $\displaystyle \frac{\sqrt{352}}{\sqrt{22}} = \sqrt{\frac{352}{22}}$

$= \sqrt{16} = \sqrt{4 \times 4} = 4$

Answer: (i) $6$ (ii) $4$

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

So $\displaystyle 2^{\frac{1}{5}} = \sqrt[5]{2}$

Answer: $\sqrt[5]{2}$

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

$= 9 \times 19 = 171$

Answer: $171$

(i) $(1 + \sqrt{7})(4 + \sqrt{7}) = 4 + \sqrt{7} + 4\sqrt{7} + 7$

$= 11 + 5\sqrt{7}$

(ii) $(1 + \sqrt{13})(2 - \sqrt{13}) = 2 - \sqrt{13} + 2\sqrt{13} - 13$

$= \sqrt{13} - 11$

Answer: (i) $11 + 5\sqrt{7}$ (ii) $\sqrt{13} - 11$

$(\sqrt{15} + \sqrt{13})(\sqrt{15} - \sqrt{13}) = (\sqrt{15})^2 - (\sqrt{13})^2$

$= 15 - 13 = 2$, a rational number.

So the rationalising factor is $\sqrt{15} - \sqrt{13}$ and the product is $2$.

Answer: Rationalising factor $\sqrt{15} - \sqrt{13}$; product $2$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{14}$, $y = \sqrt[3]{12}$.

Rationalising factor: $(\sqrt[3]{14})^2 + \sqrt[3]{14} \cdot \sqrt[3]{12} + (\sqrt[3]{12})^2$

Check: the product is $2$, a rational number.

Answer: $(\sqrt[3]{14})^2 + \sqrt[3]{14} \cdot \sqrt[3]{12} + (\sqrt[3]{12})^2$

$\sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6}$

$\sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6}$

$4\sqrt{96} = 4 \times 4\sqrt{6} = 16\sqrt{6}$

$\sqrt{294} + \sqrt{216} + 4\sqrt{96} = 7\sqrt{6} + 6\sqrt{6} + 16\sqrt{6} = 29\sqrt{6}$

So $k = 29$

Answer: $k = 29$

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

Answer: $14\sqrt{14}$

$(5\sqrt{2} + \sqrt{3})^2 = (5\sqrt{2})^2 + (\sqrt{3})^2 + 2(5\sqrt{2})(\sqrt{3})$

$(5\sqrt{2})^2 = 25 \times 2 = 50$

$2(5\sqrt{2})(\sqrt{3}) = 10\sqrt{6}$

$= 50 + 3 + 10\sqrt{6} = 53 + 10\sqrt{6}$

Answer: $53 + 10\sqrt{6}$

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

$= 16^2 - (\sqrt{2})^2 = 256 - 2 = 254$

Answer: $254$

(i) $\displaystyle \frac{9}{3 - \sqrt{6}} = \frac{9(3 + \sqrt{6})}{(3 - \sqrt{6})(3 + \sqrt{6})}$

$(3 - \sqrt{6})(3 + \sqrt{6}) = 9 - 6 = 3$

$\displaystyle = \frac{9(3 + \sqrt{6})}{3}$

$= 9 + 3\sqrt{6}$

(ii) $\displaystyle \frac{4}{\sqrt{19} + \sqrt{17}} = \frac{4(\sqrt{19} - \sqrt{17})}{(\sqrt{19} + \sqrt{17})(\sqrt{19} - \sqrt{17})}$

$(\sqrt{19} + \sqrt{17})(\sqrt{19} - \sqrt{17}) = 19 - 17 = 2$

$\displaystyle = \frac{4(\sqrt{19} - \sqrt{17})}{2}$

$= 2\sqrt{19} - 2\sqrt{17}$

Answer: (i) $9 + 3\sqrt{6}$ (ii) $2\sqrt{19} - 2\sqrt{17}$

$\displaystyle \frac{5 + \sqrt{5}}{5 - 2\sqrt{5}} = \frac{(5 + \sqrt{5})(5 + 2\sqrt{5})}{(5 - 2\sqrt{5})(5 + 2\sqrt{5})}$

Numerator: $25 + 10\sqrt{5} + 5\sqrt{5} + 10 = 35 + 15\sqrt{5}$

Denominator: $5^2 - (2\sqrt{5})^2 = 25 - 20 = 5$

$\displaystyle \frac{35 + 15\sqrt{5}}{5} = 7 + 3\sqrt{5}$

Comparing with $a + b\sqrt{5}$: $a = 7$, $b = 3$

Answer: $a = 7$, $b = 3$

$\displaystyle = \frac{(10 + \sqrt{11})^2}{(10 - \sqrt{11})(10 + \sqrt{11})} + \frac{(10 - \sqrt{11})^2}{(10 + \sqrt{11})(10 - \sqrt{11})}$

$(10 + \sqrt{11})^2 = 100 + 11 + 20\sqrt{11} = 111 + 20\sqrt{11}$, $(10 - \sqrt{11})^2 = 111 - 20\sqrt{11}$, $(10 + \sqrt{11})(10 - \sqrt{11}) = 89$

$\displaystyle = \frac{111 + 20\sqrt{11}}{89} + \frac{111 - 20\sqrt{11}}{89}$

$\displaystyle = \frac{111 + 20\sqrt{11} + 111 - 20\sqrt{11}}{89} = \frac{222}{89}$

Answer: $\displaystyle \frac{222}{89}$

$\displaystyle \frac{12}{\sqrt{3} - \sqrt{15}} = \frac{12(\sqrt{3} + \sqrt{15})}{(\sqrt{3} - \sqrt{15})(\sqrt{3} + \sqrt{15})} = \frac{12(\sqrt{3} + \sqrt{15})}{-12} = -\sqrt{15} - \sqrt{3}$

$\displaystyle \frac{4}{\sqrt{7} + \sqrt{3}} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$

$\displaystyle \frac{8}{\sqrt{15} - \sqrt{7}} = \frac{8(\sqrt{15} + \sqrt{7})}{(\sqrt{15} - \sqrt{7})(\sqrt{15} + \sqrt{7})} = \frac{8(\sqrt{15} + \sqrt{7})}{8} = \sqrt{15} + \sqrt{7}$

Sum $= (-\sqrt{15} - \sqrt{3}) - (\sqrt{7} - \sqrt{3}) + (\sqrt{15} + \sqrt{7})$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{8 - 3\sqrt{7}} = \frac{8 + 3\sqrt{7}}{(8 - 3\sqrt{7})(8 + 3\sqrt{7})}$

$\displaystyle = \frac{8 + 3\sqrt{7}}{64 - 63} = 8 + 3\sqrt{7}$

$\displaystyle \frac{1}{x} = 8 - 3\sqrt{7}$

$\displaystyle x - \frac{1}{x} = (8 + 3\sqrt{7}) - (8 - 3\sqrt{7}) = 6\sqrt{7}$

Answer: $6\sqrt{7}$

(i) $\sqrt{8 + 2\sqrt{15}} = \sqrt{5 + 3 + 2\sqrt{5 \times 3}}$

$= \sqrt{(\sqrt{5})^2 + (\sqrt{3})^2 + 2(\sqrt{5})(\sqrt{3})}$

$= \sqrt{(\sqrt{5} + \sqrt{3})^2} = \sqrt{5} + \sqrt{3}$

(ii) $6\sqrt{7} = 2 \times 3\sqrt{7} = 2\sqrt{63}$

$\sqrt{16 - 6\sqrt{7}} = \sqrt{9 + 7 - 2\sqrt{9 \times 7}}$

$= \sqrt{(3)^2 + (\sqrt{7})^2 - 2(3)(\sqrt{7})}$

$= \sqrt{(3 - \sqrt{7})^2} = 3 - \sqrt{7}$

($3 > \sqrt{7}$, so (ii) is positive.)

Answer: (i) $\sqrt{5} + \sqrt{3}$ (ii) $3 - \sqrt{7}$