Question paper, Set 5

$1^6 = 1$ and $2^6 = 64$, so $9$ is not a perfect sixth power.

So $\sqrt[6]{9}$ is not a rational number; it is a surd.

Answer: Yes. $9$ is not a perfect sixth power, so $\sqrt[6]{9}$ is not rational.

$3^3 = 27$ and $4^3 = 64$, so $50$ is not a perfect cube.

So $\sqrt[3]{50}$ is a surd, and every surd is an irrational number.

Answer: True

Change the sign of the surd part only.

(i) The conjugate of $18 - \sqrt{29}$ is $18 + \sqrt{29}$

(ii) The conjugate of $\sqrt{11} + 16$ is $-\sqrt{11} + 16$

Answer: (i) $18 + \sqrt{29}$ (ii) $-\sqrt{11} + 16$

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

$= \sqrt{70}$

Answer: $\sqrt{70}$

$\displaystyle \frac{\sqrt{10}}{\sqrt{7}} = \sqrt{\frac{10}{7}}$ (divide the numbers under the roots)

Answer: $\displaystyle \sqrt{\frac{10}{7}}$

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

(i) $\displaystyle \sqrt[4]{7} = 7^{\frac{1}{4}}$

(ii) $\displaystyle \sqrt[3]{12} = 12^{\frac{1}{3}}$

Answer: (i) $\displaystyle 7^{\frac{1}{4}}$ (ii) $\displaystyle 12^{\frac{1}{3}}$

$\sqrt[4]{125} \times \sqrt[4]{5} = \sqrt[4]{125 \times 5}$

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

Answer: $5$

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

$= 25 \times 7 = 175$

Answer: $175$

(i) $(9 + \sqrt{2})(5 - \sqrt{2}) = 45 - 9\sqrt{2} + 5\sqrt{2} - 2$

$= 43 - 4\sqrt{2}$

(ii) $(3 - \sqrt{3})(4 - \sqrt{3}) = 12 - 3\sqrt{3} - 4\sqrt{3} + 3$

$= 15 - 7\sqrt{3}$

Answer: (i) $43 - 4\sqrt{2}$ (ii) $15 - 7\sqrt{3}$

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

$= 6 - 3 = 3$, a rational number.

So the rationalising factor is $\sqrt{6} - \sqrt{3}$ and the product is $3$.

Answer: Rationalising factor $\sqrt{6} - \sqrt{3}$; product $3$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{12}$, $y = \sqrt[3]{6}$.

Rationalising factor: $(\sqrt[3]{12})^2 - \sqrt[3]{12} \cdot \sqrt[3]{6} + (\sqrt[3]{6})^2$

Check: the product is $18$, a rational number.

Answer: $(\sqrt[3]{12})^2 - \sqrt[3]{12} \cdot \sqrt[3]{6} + (\sqrt[3]{6})^2$

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

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

$5\sqrt{96} = 5 \times 4\sqrt{6} = 20\sqrt{6}$

$\sqrt{216} + \sqrt{24} + 5\sqrt{96} = 6\sqrt{6} + 2\sqrt{6} + 20\sqrt{6} = 28\sqrt{6}$

So $k = 28$

Answer: $k = 28$

$(\sqrt{7})^4 = [(\sqrt{7})^2]^2 = 7^2 = 49$

Answer: $49$

$(\sqrt{11} - \sqrt{6})^2 = (\sqrt{11})^2 + (\sqrt{6})^2 - 2(\sqrt{11})(\sqrt{6})$

$= 11 + 6 - 2\sqrt{66} = 17 - 2\sqrt{66}$

Answer: $17 - 2\sqrt{66}$

$(7 + \sqrt{5})(9 + 3\sqrt{5}) = 63 + 21\sqrt{5} + 9\sqrt{5} + 15$

[$\sqrt{5} \times 3\sqrt{5} = 3 \times 5 = 15$]

$= 78 + 30\sqrt{5}$

Answer: $78 + 30\sqrt{5}$

$\displaystyle \frac{5 + \sqrt{14}}{5 - \sqrt{14}} - \frac{5 - \sqrt{14}}{5 + \sqrt{14}} = \frac{(5 + \sqrt{14})^2 - (5 - \sqrt{14})^2}{(5 - \sqrt{14})(5 + \sqrt{14})}$

$(5 + \sqrt{14})^2 = 39 + 10\sqrt{14}$

$(5 - \sqrt{14})^2 = 39 - 10\sqrt{14}$

$(5 + \sqrt{14})(5 - \sqrt{14}) = 25 - 14 = 11$

$\displaystyle = \frac{(39 + 10\sqrt{14}) - (39 - 10\sqrt{14})}{11} = \frac{20\sqrt{14}}{11}$

Answer: $\displaystyle \frac{20\sqrt{14}}{11}$

$\displaystyle \frac{11}{2 - \sqrt{15}} = \frac{11(2 + \sqrt{15})}{(2 - \sqrt{15})(2 + \sqrt{15})} = \frac{11(2 + \sqrt{15})}{-11} = -2 - \sqrt{15}$

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

$\displaystyle \frac{1}{\sqrt{5} + 2} = \frac{\sqrt{5} - 2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \sqrt{5} - 2$

Sum $= (-2 - \sqrt{15}) + (\sqrt{15} + \sqrt{5}) - (\sqrt{5} - 2)$

$= 0$ (all the terms cancel)

Answer: $0$

$\displaystyle x = \frac{1}{15 - 4\sqrt{14}} = \frac{15 + 4\sqrt{14}}{(15 - 4\sqrt{14})(15 + 4\sqrt{14})}$

$\displaystyle = \frac{15 + 4\sqrt{14}}{225 - 224} = 15 + 4\sqrt{14}$

$x - 15 = 4\sqrt{14}$

$(x - 15)^2 = (4\sqrt{14})^2 \Rightarrow x^2 - 30x + 225 = 224$

$\Rightarrow x^2 - 30x + 1 = 0$

Dividing $x^3 - 32x^2 + 61x$ by $x^2 - 30x + 1$: quotient $x - 2$, remainder $2$

$x^3 - 32x^2 + 61x = (x^2 - 30x + 1)(x - 2) + 2$

$= 0 \times (x - 2) + 2 = 2$

Answer: $2$

$\sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$

$\sqrt{10 - \sqrt{84}} = \sqrt{7 + 3 - 2\sqrt{7 \times 3}}$

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

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

($\sqrt{7} > \sqrt{3}$, so this is the positive square root.)

Answer: $\sqrt{7} - \sqrt{3}$

$\displaystyle \sqrt{15 - \sqrt{221}} = \sqrt{\frac{2(15 - \sqrt{221})}{2}} = \frac{\sqrt{30 - 2\sqrt{221}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{17 + 13 - 2\sqrt{17 \times 13}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{17})^2 + (\sqrt{13})^2 - 2(\sqrt{17})(\sqrt{13})}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{17} - \sqrt{13})^2}}{\sqrt{2}} = \frac{\sqrt{17} - \sqrt{13}}{\sqrt{2}}$

$\displaystyle \frac{\sqrt{17} - \sqrt{13}}{\sqrt{2}} = \frac{(\sqrt{17} - \sqrt{13}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{34} - \sqrt{26}}{2}$

Answer: $\displaystyle \frac{\sqrt{34} - \sqrt{26}}{2}$