Question paper, Set 6

$\sqrt[4]{625} = \sqrt[4]{5 \times 5 \times 5 \times 5} = 5$

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

Answer: No. $\sqrt[4]{625} = 5$, a rational number.

$7^2 = 49$ and $8^2 = 64$, so $58$ is not a perfect square.

So $\sqrt{58}$ is a surd, and $3\sqrt{58}$ is in the form $a \cdot \sqrt{b}$: a mixed surd.

Answer: Yes. It is $3$ times the surd $\sqrt{58}$.

Change the sign of the surd part only.

(i) The conjugate of $14 - \sqrt{14}$ is $14 + \sqrt{14}$

(ii) The conjugate of $10 + \sqrt{11}$ is $10 - \sqrt{11}$

Answer: (i) $14 + \sqrt{14}$ (ii) $10 - \sqrt{11}$

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

$= \sqrt{17^2} = 17$

Answer: $17$

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

(i) $\displaystyle \sqrt[5]{23} = 23^{\frac{1}{5}}$

(ii) $\displaystyle \sqrt[4]{14} = 14^{\frac{1}{4}}$

Answer: (i) $\displaystyle 23^{\frac{1}{5}}$ (ii) $\displaystyle 14^{\frac{1}{4}}$

(i) $\sqrt[5]{9} \times \sqrt[5]{27} = \sqrt[5]{9 \times 27}$

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

(ii) $\displaystyle \frac{\sqrt[3]{80}}{\sqrt[3]{10}} = \sqrt[3]{\frac{80}{10}}$

$= \sqrt[3]{8} = \sqrt[3]{2^3} = 2$

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

$(10\sqrt{23})^2 = 10^2 \times (\sqrt{23})^2$

$= 100 \times 23 = 2300$

Answer: $2300$

$(4 + \sqrt{6})(1 - \sqrt{6}) = 4 - 4\sqrt{6} + \sqrt{6} - 6$

$= -2 - 3\sqrt{6}$

Answer: $-2 - 3\sqrt{6}$

$\sqrt{21} \times \sqrt{21} = 1 \times (\sqrt{21})^2 = 1 \times 21 = 21$, a rational number.

So $\sqrt{21}$ is a rationalising factor of $\sqrt{21}$.

Answer: $\sqrt{21}$ (since $\sqrt{21} \times \sqrt{21} = 21$)

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

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

$\sqrt{150} + \sqrt{294} = 5\sqrt{6} + 7\sqrt{6} = 12\sqrt{6}$

Answer: $12\sqrt{6}$

(i) $(\sqrt{5})^{7} = [(\sqrt{5})^2]^{3} \times \sqrt{5} = 5^{3} \times \sqrt{5} = 125\sqrt{5}$

(ii) $(8\sqrt{2})^{3} = 8^{3} \times (\sqrt{2})^{3} = 512 \times 2\sqrt{2} = 1024\sqrt{2}$

Answer: (i) $125\sqrt{5}$ (ii) $1024\sqrt{2}$

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

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

$= 121 + 11 - 22\sqrt{11} = 132 - 22\sqrt{11}$

Answer: $132 - 22\sqrt{11}$

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

$= 18^2 - (\sqrt{30})^2 = 324 - 30 = 294$

Answer: $294$

$\displaystyle \frac{1}{\sqrt{15} - \sqrt{13}} = \frac{\sqrt{15} + \sqrt{13}}{(\sqrt{15} - \sqrt{13})(\sqrt{15} + \sqrt{13})}$

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

$\displaystyle = \frac{\sqrt{15} + \sqrt{13}}{2}$

Answer: $\displaystyle \frac{\sqrt{15} + \sqrt{13}}{2}$

$\displaystyle \frac{8 + 2\sqrt{2}}{4 - 2\sqrt{2}} = \frac{(8 + 2\sqrt{2})(4 + 2\sqrt{2})}{(4 - 2\sqrt{2})(4 + 2\sqrt{2})}$

Numerator: $32 + 16\sqrt{2} + 8\sqrt{2} + 8 = 40 + 24\sqrt{2}$

Denominator: $4^2 - (2\sqrt{2})^2 = 16 - 8 = 8$

$\displaystyle \frac{40 + 24\sqrt{2}}{8} = 5 + 3\sqrt{2}$

Comparing with $a + b\sqrt{2}$: $a = 5$, $b = 3$

Answer: $a = 5$, $b = 3$

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

$(6 + \sqrt{10})^2 = 36 + 10 + 12\sqrt{10} = 46 + 12\sqrt{10}$, $(6 - \sqrt{10})^2 = 46 - 12\sqrt{10}$, $(6 + \sqrt{10})(6 - \sqrt{10}) = 26$

$\displaystyle = \frac{46 + 12\sqrt{10}}{26} + \frac{46 - 12\sqrt{10}}{26}$

$\displaystyle = \frac{46 + 12\sqrt{10} + 46 - 12\sqrt{10}}{26} = \frac{46}{13}$

Answer: $\displaystyle \frac{46}{13}$

$\displaystyle \frac{5}{\sqrt{14} + 3} = \frac{5(\sqrt{14} - 3)}{(\sqrt{14} + 3)(\sqrt{14} - 3)} = \frac{5(\sqrt{14} - 3)}{5} = \sqrt{14} - 3$

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

$\displaystyle \frac{4}{\sqrt{5} - 3} = \frac{4(\sqrt{5} + 3)}{(\sqrt{5} - 3)(\sqrt{5} + 3)} = \frac{4(\sqrt{5} + 3)}{-4} = -3 - \sqrt{5}$

Sum $= (\sqrt{14} - 3) - (\sqrt{14} + \sqrt{5}) - (-3 - \sqrt{5})$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{11 + 2\sqrt{30}} = \frac{11 - 2\sqrt{30}}{(11 + 2\sqrt{30})(11 - 2\sqrt{30})}$

$\displaystyle = \frac{11 - 2\sqrt{30}}{121 - 120} = 11 - 2\sqrt{30}$

$x - 11 = -2\sqrt{30}$

$(x - 11)^2 = (-2\sqrt{30})^2 \Rightarrow x^2 - 22x + 121 = 120$

$\Rightarrow x^2 - 22x + 1 = 0$

Dividing $x^3 - 21x^2 - 21x + 10$ by $x^2 - 22x + 1$: quotient $x + 1$, remainder $9$

$x^3 - 21x^2 - 21x + 10 = (x^2 - 22x + 1)(x + 1) + 9$

$= 0 \times (x + 1) + 9 = 9$

Answer: $9$

$\sqrt{112} = \sqrt{4 \times 28} = 2\sqrt{28}$

$\sqrt{11 + \sqrt{112}} = \sqrt{7 + 4 + 2\sqrt{7 \times 4}}$

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

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

Answer: $\sqrt{7} + 2$

$\displaystyle \sqrt{11 + \sqrt{85}} = \sqrt{\frac{2(11 + \sqrt{85})}{2}} = \frac{\sqrt{22 + 2\sqrt{85}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{17 + 5 + 2\sqrt{17 \times 5}}}{\sqrt{2}}$

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

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

$\displaystyle \frac{\sqrt{17} + \sqrt{5}}{\sqrt{2}} = \frac{(\sqrt{17} + \sqrt{5}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{34} + \sqrt{10}}{2}$

Answer: $\displaystyle \frac{\sqrt{34} + \sqrt{10}}{2}$