Question paper, Set 7

(i) $\sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = 10$, rational, not a surd

(ii) $1^3 = 1$ and $2^3 = 8$, so $3$ is not a perfect cube. So $\sqrt[3]{3}$ is a surd.

(iii) $1^5 = 1$ and $2^5 = 32$, so $17$ is not a perfect fifth power. So $\sqrt[5]{17}$ is a surd.

(iv) $\sqrt[6]{729} = \sqrt[6]{3 \times 3 \times 3 \times 3 \times 3 \times 3} = 3$, rational, not a surd

Surds: (ii) and (iii)

Answer: (ii) and (iii)

$\pi$ and $e$ are irrational numbers but they are not surds.

Answer: False

Conjugate surds: $a + \sqrt{b}$ and $a - \sqrt{b}$ differ only in the sign of the surd.

The conjugate of $\sqrt{5} + 1$ is $-\sqrt{5} + 1$.

Answer: $-\sqrt{5} + 1$

$\displaystyle \frac{\sqrt{22}}{\sqrt{21}} = \sqrt{\frac{22}{21}}$ (divide the numbers under the roots)

Answer: $\displaystyle \sqrt{\frac{22}{21}}$

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

So $\displaystyle \sqrt[6]{11} = 11^{\frac{1}{6}}$

Answer: $\displaystyle 11^{\frac{1}{6}}$

$\displaystyle \frac{\sqrt[3]{320}}{\sqrt[3]{5}} = \sqrt[3]{\frac{320}{5}}$

$= \sqrt[3]{64} = \sqrt[3]{4^3} = 4$

Answer: $4$

(i) $(8\sqrt{6})^2 = 8^2 \times (\sqrt{6})^2$

$= 64 \times 6 = 384$

(ii) $(4\sqrt{5})^2 = 4^2 \times (\sqrt{5})^2$

$= 16 \times 5 = 80$

Answer: (i) $384$ (ii) $80$

(i) $(9 - \sqrt{10})(7 + \sqrt{10}) = 63 + 9\sqrt{10} - 7\sqrt{10} - 10$

$= 53 + 2\sqrt{10}$

(ii) $(9 + \sqrt{11})(4 - \sqrt{11}) = 36 - 9\sqrt{11} + 4\sqrt{11} - 11$

$= 25 - 5\sqrt{11}$

Answer: (i) $53 + 2\sqrt{10}$ (ii) $25 - 5\sqrt{11}$

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

$= 10 - 6 = 4$, a rational number.

So the rationalising factor is $\sqrt{10} - \sqrt{6}$ and the product is $4$.

Answer: Rationalising factor $\sqrt{10} - \sqrt{6}$; product $4$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{9}$, $y = \sqrt[3]{2}$.

Rationalising factor: $(\sqrt[3]{9})^2 - \sqrt[3]{9} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^2$

$= \sqrt[3]{81} - \sqrt[3]{18} + \sqrt[3]{4}$

Product $= (\sqrt[3]{9})^3 + (\sqrt[3]{2})^3 = 9 + 2 = 11$

Answer: $(\sqrt[3]{9})^2 - \sqrt[3]{9} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^2$, that is $\sqrt[3]{81} - \sqrt[3]{18} + \sqrt[3]{4}$; product $11$

$\sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7}$

$\sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7}$

$\sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7}$

$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$

$3\sqrt{252} + \sqrt{175} - 3\sqrt{112} + \sqrt{63} = 18\sqrt{7} + 5\sqrt{7} - 12\sqrt{7} + 3\sqrt{7}$

$= 14\sqrt{7}$

Answer: $14\sqrt{7}$

$(\sqrt{10})^4 = [(\sqrt{10})^2]^2 = 10^2 = 100$

Answer: $100$

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

$= 13 + 2 + 2\sqrt{26} = 15 + 2\sqrt{26}$

Answer: $15 + 2\sqrt{26}$

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

$= (\sqrt{14})^2 - (\sqrt{5})^2 = 14 - 5 = 9$

Answer: $9$

$\displaystyle \frac{1}{\sqrt{11}} = \frac{1 \times \sqrt{11}}{\sqrt{11} \times \sqrt{11}} = \frac{\sqrt{11}}{11}$

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

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

Numerator: $3 + 2\sqrt{2} + 3\sqrt{2} + 4 = 7 + 5\sqrt{2}$

Denominator: $3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$

So it equals $7 + 5\sqrt{2}$

Comparing with $a + b\sqrt{2}$: $a = 7$, $b = 5$

Answer: $a = 7$, $b = 5$

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

$(\sqrt{15} + \sqrt{13})^2 = 28 + 2\sqrt{195}$

$(\sqrt{15} - \sqrt{13})^2 = 28 - 2\sqrt{195}$

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

$\displaystyle = \frac{28 + 2\sqrt{195}}{2} + \frac{28 - 2\sqrt{195}}{2}$

$\displaystyle = \frac{28 + 2\sqrt{195} + 28 - 2\sqrt{195}}{2} = 28$

Answer: $28$

$\displaystyle x = \frac{1}{5 - 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})}$

$\displaystyle = \frac{5 + 2\sqrt{6}}{25 - 24} = 5 + 2\sqrt{6}$

$x - 5 = 2\sqrt{6}$

$(x - 5)^2 = (2\sqrt{6})^2 \Rightarrow x^2 - 10x + 25 = 24$

$\Rightarrow x^2 - 10x + 1 = 0$

Dividing $x^3 - 13x^2 + 31x + 1$ by $x^2 - 10x + 1$: quotient $x - 3$, remainder $4$

$x^3 - 13x^2 + 31x + 1 = (x^2 - 10x + 1)(x - 3) + 4$

$= 0 \times (x - 3) + 4 = 4$

Answer: $4$

(i) $\sqrt{120} = \sqrt{4 \times 30} = 2\sqrt{30}$

$\sqrt{13 + \sqrt{120}} = \sqrt{10 + 3 + 2\sqrt{10 \times 3}}$

$= \sqrt{(\sqrt{10})^2 + (\sqrt{3})^2 + 2(\sqrt{10})(\sqrt{3})}$

$= \sqrt{(\sqrt{10} + \sqrt{3})^2} = \sqrt{10} + \sqrt{3}$

(ii) $\sqrt{20 - 2\sqrt{84}} = \sqrt{14 + 6 - 2\sqrt{14 \times 6}}$

$= \sqrt{(\sqrt{14})^2 + (\sqrt{6})^2 - 2(\sqrt{14})(\sqrt{6})}$

$= \sqrt{(\sqrt{14} - \sqrt{6})^2} = \sqrt{14} - \sqrt{6}$

($\sqrt{14} > \sqrt{6}$, so (ii) is positive.)

Answer: (i) $\sqrt{10} + \sqrt{3}$ (ii) $\sqrt{14} - \sqrt{6}$

$\displaystyle \sqrt{15 + \sqrt{209}} = \sqrt{\frac{2(15 + \sqrt{209})}{2}} = \frac{\sqrt{30 + 2\sqrt{209}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{19 + 11 + 2\sqrt{19 \times 11}}}{\sqrt{2}}$

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

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

Answer: $\displaystyle \frac{\sqrt{19} + \sqrt{11}}{\sqrt{2}}$