Question paper, Set 10

$\sqrt{256} = \sqrt{16 \times 16} = 16$

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

Answer: No. $\sqrt{256} = 16$, a rational number.

$1^6 = 1$ and $2^6 = 64$, so $23$ is not a perfect sixth power.

So $\sqrt[6]{23}$ is a surd, and every surd is an irrational number.

Answer: True

Change the sign of the surd part only.

(i) The conjugate of $19 - \sqrt{10}$ is $19 + \sqrt{10}$

(ii) The conjugate of $7 + \sqrt{29}$ is $7 - \sqrt{29}$

Answer: (i) $19 + \sqrt{10}$ (ii) $7 - \sqrt{29}$

$\sqrt{14} \times \sqrt{19} = \sqrt{14 \times 19}$

$= \sqrt{266}$

Answer: $\sqrt{266}$

$\displaystyle \frac{\sqrt{11}}{\sqrt{2}} = \sqrt{\frac{11}{2}}$ (divide the numbers under the roots)

Answer: $\displaystyle \sqrt{\frac{11}{2}}$

$\displaystyle 64^{\frac{1}{3}} = \sqrt[3]{64}$

$= \sqrt[3]{4^3} = \sqrt[3]{4 \times 4 \times 4} = 4$

Answer: $4$

$\sqrt[3]{32} \times \sqrt[3]{16} = \sqrt[3]{32 \times 16}$

$= \sqrt[3]{512} = \sqrt[3]{8^3} = 8$

Answer: $8$

$(7\sqrt{17})^2 = 7^2 \times (\sqrt{17})^2$

$= 49 \times 17 = 833$

Answer: $833$

(i) $(7 - \sqrt{7})(2 + \sqrt{7}) = 14 + 7\sqrt{7} - 2\sqrt{7} - 7$

$= 7 + 5\sqrt{7}$

(ii) $(8 - \sqrt{11})(5 - \sqrt{11}) = 40 - 8\sqrt{11} - 5\sqrt{11} + 11$

$= 51 - 13\sqrt{11}$

Answer: (i) $7 + 5\sqrt{7}$ (ii) $51 - 13\sqrt{11}$

$(\sqrt{19} - \sqrt{11})(\sqrt{19} + \sqrt{11}) = (\sqrt{19})^2 - (\sqrt{11})^2$

$= 19 - 11 = 8$, a rational number.

So the rationalising factor is $\sqrt{19} + \sqrt{11}$ and the product is $8$.

Answer: Rationalising factor $\sqrt{19} + \sqrt{11}$; product $8$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{10}$, $y = \sqrt[3]{6}$.

Rationalising factor: $(\sqrt[3]{10})^2 + \sqrt[3]{10} \cdot \sqrt[3]{6} + (\sqrt[3]{6})^2$

Check: the product is $4$, a rational number.

Answer: $(\sqrt[3]{10})^2 + \sqrt[3]{10} \cdot \sqrt[3]{6} + (\sqrt[3]{6})^2$

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

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

$3\sqrt{294} = 3 \times 7\sqrt{6} = 21\sqrt{6}$

$\sqrt{24} + \sqrt{96} + 3\sqrt{294} = 2\sqrt{6} + 4\sqrt{6} + 21\sqrt{6} = 27\sqrt{6}$

So $k = 27$

Answer: $k = 27$

$(\sqrt{13})^4 = [(\sqrt{13})^2]^2 = 13^2 = 169$

Answer: $169$

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

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

$= 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$

Answer: $3 + 2\sqrt{2}$

$(2 + 2\sqrt{5})(4 - \sqrt{5}) = 8 - 2\sqrt{5} + 8\sqrt{5} - 10$

[$2\sqrt{5} \times (-\sqrt{5}) = -2 \times 5 = -10$]

$= 6\sqrt{5} - 2$

Answer: $6\sqrt{5} - 2$

$\displaystyle = \frac{(9 + 4\sqrt{5})^2}{(9 - 4\sqrt{5})(9 + 4\sqrt{5})} + \frac{(9 - 4\sqrt{5})^2}{(9 + 4\sqrt{5})(9 - 4\sqrt{5})}$

$(9 + 4\sqrt{5})^2 = 161 + 72\sqrt{5}$

$(9 - 4\sqrt{5})^2 = 161 - 72\sqrt{5}$

$(9 + 4\sqrt{5})(9 - 4\sqrt{5}) = 1$

$\displaystyle = \frac{161 + 72\sqrt{5}}{1} + \frac{161 - 72\sqrt{5}}{1}$

$\displaystyle = \frac{161 + 72\sqrt{5} + 161 - 72\sqrt{5}}{1} = 322$

Answer: $322$

$\displaystyle \frac{9}{\sqrt{6} - \sqrt{15}} = \frac{9(\sqrt{6} + \sqrt{15})}{(\sqrt{6} - \sqrt{15})(\sqrt{6} + \sqrt{15})} = \frac{9(\sqrt{6} + \sqrt{15})}{-9} = -\sqrt{15} - \sqrt{6}$

$\displaystyle \frac{10}{\sqrt{15} + \sqrt{5}} = \frac{10(\sqrt{15} - \sqrt{5})}{(\sqrt{15} + \sqrt{5})(\sqrt{15} - \sqrt{5})} = \frac{10(\sqrt{15} - \sqrt{5})}{10} = \sqrt{15} - \sqrt{5}$

$\displaystyle \frac{1}{\sqrt{5} - \sqrt{6}} = \frac{\sqrt{5} + \sqrt{6}}{(\sqrt{5} - \sqrt{6})(\sqrt{5} + \sqrt{6})} = \frac{\sqrt{5} + \sqrt{6}}{-1} = -\sqrt{6} - \sqrt{5}$

Sum $= (-\sqrt{15} - \sqrt{6}) + (\sqrt{15} - \sqrt{5}) - (-\sqrt{6} - \sqrt{5})$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{10 + 3\sqrt{11}} = \frac{10 - 3\sqrt{11}}{(10 + 3\sqrt{11})(10 - 3\sqrt{11})}$

$\displaystyle = \frac{10 - 3\sqrt{11}}{100 - 99} = 10 - 3\sqrt{11}$

$x - 10 = -3\sqrt{11}$

$(x - 10)^2 = (-3\sqrt{11})^2 \Rightarrow x^2 - 20x + 100 = 99$

$\Rightarrow x^2 - 20x + 1 = 0$

Dividing $x^3 - 17x^2 - 59x + 12$ by $x^2 - 20x + 1$: quotient $x + 3$, remainder $9$

$x^3 - 17x^2 - 59x + 12 = (x^2 - 20x + 1)(x + 3) + 9$

$= 0 \times (x + 3) + 9 = 9$

Answer: $9$

$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$

$\sqrt{8 - \sqrt{28}} = \sqrt{7 + 1 - 2\sqrt{7 \times 1}}$

$= \sqrt{(\sqrt{7})^2 + (1)^2 - 2(\sqrt{7})(1)}$

$= \sqrt{(\sqrt{7} - 1)^2} = \sqrt{7} - 1$

($\sqrt{7} > 1$, so this is the positive square root.)

Answer: $\sqrt{7} - 1$

$\displaystyle \sqrt{11 - \sqrt{57}} = \sqrt{\frac{2(11 - \sqrt{57})}{2}} = \frac{\sqrt{22 - 2\sqrt{57}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{19 + 3 - 2\sqrt{19 \times 3}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{19})^2 + (\sqrt{3})^2 - 2(\sqrt{19})(\sqrt{3})}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{19} - \sqrt{3})^2}}{\sqrt{2}} = \frac{\sqrt{19} - \sqrt{3}}{\sqrt{2}}$

$\displaystyle \frac{\sqrt{19} - \sqrt{3}}{\sqrt{2}} = \frac{(\sqrt{19} - \sqrt{3}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{38} - \sqrt{6}}{2}$

Answer: $\displaystyle \frac{\sqrt{38} - \sqrt{6}}{2}$