Question paper, Set 1

$1^6 = 1$ and $2^6 = 64$, so $10$ is not a perfect sixth power.

So $\sqrt[6]{10}$ is not a rational number; it is a surd.

Answer: Yes. $10$ is not a perfect sixth power, so $\sqrt[6]{10}$ is not rational.

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

$2\sqrt{49} = 2 \times 7 = 14$, which is rational, so it is not a surd.

Answer: No. $2\sqrt{49} = 14$, a rational number.

Change the sign of the surd part only.

(i) The conjugate of $9 - \sqrt{5}$ is $9 + \sqrt{5}$

(ii) The conjugate of $\sqrt{15} + 5$ is $-\sqrt{15} + 5$

Answer: (i) $9 + \sqrt{5}$ (ii) $-\sqrt{15} + 5$

$\sqrt{38} \times \sqrt{38} = (\sqrt{38})^2$

$= \sqrt{38^2} = 38$

Answer: $38$

$\displaystyle 625^{\frac{1}{2}} = \sqrt{625}$

$= \sqrt{25^2} = \sqrt{25 \times 25} = 25$

Answer: $25$

(i) $\sqrt[6]{16} \times \sqrt[6]{4} = \sqrt[6]{16 \times 4}$

$= \sqrt[6]{64} = \sqrt[6]{2^6} = 2$

(ii) $\displaystyle \frac{\sqrt[3]{243}}{\sqrt[3]{9}} = \sqrt[3]{\frac{243}{9}}$

$= \sqrt[3]{27} = \sqrt[3]{3^3} = 3$

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

$(12\sqrt{3})^2 = 12^2 \times (\sqrt{3})^2$

$= 144 \times 3 = 432$

Answer: $432$

$(1 + \sqrt{11})(3 + \sqrt{11}) = 3 + \sqrt{11} + 3\sqrt{11} + 11$

$= 14 + 4\sqrt{11}$

Answer: $14 + 4\sqrt{11}$

$(11 - \sqrt{26})(11 + \sqrt{26}) = 11^2 - (\sqrt{26})^2 = 121 - 26 = 95$, a rational number.

So the rationalising factor is $11 + \sqrt{26}$.

Answer: $11 + \sqrt{26}$

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

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

$\sqrt{216} - \sqrt{54} = 6\sqrt{6} - 3\sqrt{6} = 3\sqrt{6}$

Answer: $3\sqrt{6}$

(i) $(\sqrt{6})^{5} = [(\sqrt{6})^2]^{2} \times \sqrt{6} = 6^{2} \times \sqrt{6} = 36\sqrt{6}$

(ii) $(4\sqrt{10})^{3} = 4^{3} \times (\sqrt{10})^{3} = 64 \times 10\sqrt{10} = 640\sqrt{10}$

Answer: (i) $36\sqrt{6}$ (ii) $640\sqrt{10}$

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

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

$= 100 + 11 + 20\sqrt{11} = 111 + 20\sqrt{11}$

Answer: $111 + 20\sqrt{11}$

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

$= 19^2 - (\sqrt{19})^2 = 361 - 19 = 342$

Answer: $342$

$\displaystyle \frac{1}{4 + \sqrt{5}} = \frac{4 - \sqrt{5}}{(4 + \sqrt{5})(4 - \sqrt{5})}$

$(4 + \sqrt{5})(4 - \sqrt{5}) = 4^2 - (\sqrt{5})^2 = 16 - 5 = 11$

$\displaystyle = \frac{4 - \sqrt{5}}{11}$

Answer: $\displaystyle \frac{4 - \sqrt{5}}{11}$

$\displaystyle \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{(\sqrt{6} + \sqrt{5})^2}{(\sqrt{6} - \sqrt{5})(\sqrt{6} + \sqrt{5})}$

$\displaystyle = \frac{6 + 5 + 2\sqrt{30}}{6 - 5}$

$= 11 + 2\sqrt{30}$

Comparing with $a + b\sqrt{30}$: $a = 11$, $b = 2$

Answer: $a = 11$, $b = 2$

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

$(3 + \sqrt{3})^2 = 9 + 3 + 6\sqrt{3} = 12 + 6\sqrt{3}$, $(3 - \sqrt{3})^2 = 12 - 6\sqrt{3}$, $(3 + \sqrt{3})(3 - \sqrt{3}) = 6$

$\displaystyle = \frac{12 + 6\sqrt{3}}{6} + \frac{12 - 6\sqrt{3}}{6}$

$\displaystyle = \frac{12 + 6\sqrt{3} + 12 - 6\sqrt{3}}{6} = 4$

Answer: $4$

$\displaystyle \frac{4}{\sqrt{10} - \sqrt{6}} = \frac{4(\sqrt{10} + \sqrt{6})}{(\sqrt{10} - \sqrt{6})(\sqrt{10} + \sqrt{6})} = \frac{4(\sqrt{10} + \sqrt{6})}{4} = \sqrt{10} + \sqrt{6}$

$\displaystyle \frac{6}{2 - \sqrt{10}} = \frac{6(2 + \sqrt{10})}{(2 - \sqrt{10})(2 + \sqrt{10})} = \frac{6(2 + \sqrt{10})}{-6} = -2 - \sqrt{10}$

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

Sum $= (\sqrt{10} + \sqrt{6}) + (-2 - \sqrt{10}) - (\sqrt{6} - 2)$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}$

$\displaystyle = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}$

$x - 3 = -2\sqrt{2}$

$(x - 3)^2 = (-2\sqrt{2})^2 \Rightarrow x^2 - 6x + 9 = 8$

$\Rightarrow x^2 - 6x + 1 = 0$

Dividing $x^3 - 7x^2 + 7x + 4$ by $x^2 - 6x + 1$: quotient $x - 1$, remainder $5$

$x^3 - 7x^2 + 7x + 4 = (x^2 - 6x + 1)(x - 1) + 5$

$= 0 \times (x - 1) + 5 = 5$

Answer: $5$

$\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

$\sqrt{7 + \sqrt{24}} = \sqrt{6 + 1 + 2\sqrt{6 \times 1}}$

$= \sqrt{(\sqrt{6})^2 + (1)^2 + 2(\sqrt{6})(1)}$

$= \sqrt{(\sqrt{6} + 1)^2} = \sqrt{6} + 1$

Answer: $\sqrt{6} + 1$

$\displaystyle \sqrt{13 - \sqrt{69}} = \sqrt{\frac{2(13 - \sqrt{69})}{2}} = \frac{\sqrt{26 - 2\sqrt{69}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{23 + 3 - 2\sqrt{23 \times 3}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23})^2 + (\sqrt{3})^2 - 2(\sqrt{23})(\sqrt{3})}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23} - \sqrt{3})^2}}{\sqrt{2}} = \frac{\sqrt{23} - \sqrt{3}}{\sqrt{2}}$

$\displaystyle \frac{\sqrt{23} - \sqrt{3}}{\sqrt{2}} = \frac{(\sqrt{23} - \sqrt{3}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{46} - \sqrt{6}}{2}$

Answer: $\displaystyle \frac{\sqrt{46} - \sqrt{6}}{2}$