Question paper, Set 2

(i) $1^5 = 1$ and $2^5 = 32$, so $27$ is not a perfect fifth power. So $\sqrt[5]{27}$ is a surd.

(ii) $\sqrt[6]{64} = \sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2} = 2$, rational, not a surd

(iii) $\sqrt{25} = \sqrt{5 \times 5} = 5$, rational, not a surd

(iv) $1^5 = 1$ and $2^5 = 32$, so $4$ is not a perfect fifth power. So $\sqrt[5]{4}$ is a surd.

Surds: (i) and (iv)

Answer: (i) and (iv)

$e = 2.71\ldots$ is a non-terminating, non-recurring decimal, so it is irrational.

But $e$ is not the $n$th root of a rational number, so it is not a surd.

Answer: False

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

The conjugate of $3 + \sqrt{30}$ is $3 - \sqrt{30}$.

Answer: $3 - \sqrt{30}$

$\displaystyle \frac{\sqrt{176}}{\sqrt{11}} = \sqrt{\frac{176}{11}}$

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

Answer: $4$

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

So $\displaystyle 25^{\frac{1}{3}} = \sqrt[3]{25}$

Answer: $\sqrt[3]{25}$

$\displaystyle \frac{\sqrt[5]{64}}{\sqrt[5]{2}} = \sqrt[5]{\frac{64}{2}}$

$= \sqrt[5]{32} = \sqrt[5]{2^5} = 2$

Answer: $2$

(i) $(6\sqrt{10})^2 = 6^2 \times (\sqrt{10})^2$

$= 36 \times 10 = 360$

(ii) $(2\sqrt{21})^2 = 2^2 \times (\sqrt{21})^2$

$= 4 \times 21 = 84$

Answer: (i) $360$ (ii) $84$

(i) $(4 - \sqrt{2})(8 + \sqrt{2}) = 32 + 4\sqrt{2} - 8\sqrt{2} - 2$

$= 30 - 4\sqrt{2}$

(ii) $(9 - \sqrt{5})(8 - \sqrt{5}) = 72 - 9\sqrt{5} - 8\sqrt{5} + 5$

$= 77 - 17\sqrt{5}$

Answer: (i) $30 - 4\sqrt{2}$ (ii) $77 - 17\sqrt{5}$

$(\sqrt{13} + \sqrt{10})(\sqrt{13} - \sqrt{10}) = (\sqrt{13})^2 - (\sqrt{10})^2$

$= 13 - 10 = 3$, a rational number.

So the rationalising factor is $\sqrt{13} - \sqrt{10}$ and the product is $3$.

Answer: Rationalising factor $\sqrt{13} - \sqrt{10}$; product $3$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{13}$, $y = \sqrt[3]{12}$.

Rationalising factor: $(\sqrt[3]{13})^2 - \sqrt[3]{13} \cdot \sqrt[3]{12} + (\sqrt[3]{12})^2$

$= \sqrt[3]{169} - \sqrt[3]{156} + \sqrt[3]{144}$

Product $= (\sqrt[3]{13})^3 + (\sqrt[3]{12})^3 = 13 + 12 = 25$

Answer: $(\sqrt[3]{13})^2 - \sqrt[3]{13} \cdot \sqrt[3]{12} + (\sqrt[3]{12})^2$, that is $\sqrt[3]{169} - \sqrt[3]{156} + \sqrt[3]{144}$; product $25$

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

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

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

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

$2\sqrt{96} + \sqrt{150} - 2\sqrt{54} + \sqrt{216} = 8\sqrt{6} + 5\sqrt{6} - 6\sqrt{6} + 6\sqrt{6}$

$= 13\sqrt{6}$

Answer: $13\sqrt{6}$

$(\sqrt{17})^4 = [(\sqrt{17})^2]^2 = 17^2 = 289$

Answer: $289$

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

$= 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}$

Answer: $12 + 2\sqrt{35}$

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

$= 4^2 - (4\sqrt{7})^2$

$(4\sqrt{7})^2 = 16 \times 7 = 112$

$= 16 - 112 = -96$

Answer: $-96$

$\displaystyle \frac{12}{\sqrt{6}} = \frac{12 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{12\sqrt{6}}{6}$

$= 2\sqrt{6}$

Answer: $2\sqrt{6}$

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

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

Denominator: $2^2 - (\sqrt{3})^2 = 4 - 3 = 1$

So it equals $5 + 3\sqrt{3}$

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

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

$\displaystyle \frac{9 + \sqrt{7}}{9 - \sqrt{7}} - \frac{9 - \sqrt{7}}{9 + \sqrt{7}} = \frac{(9 + \sqrt{7})^2 - (9 - \sqrt{7})^2}{(9 - \sqrt{7})(9 + \sqrt{7})}$

$(9 + \sqrt{7})^2 = 88 + 18\sqrt{7}$

$(9 - \sqrt{7})^2 = 88 - 18\sqrt{7}$

$(9 + \sqrt{7})(9 - \sqrt{7}) = 81 - 7 = 74$

$\displaystyle = \frac{(88 + 18\sqrt{7}) - (88 - 18\sqrt{7})}{74} = \frac{18\sqrt{7}}{37}$

Answer: $\displaystyle \frac{18\sqrt{7}}{37}$

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

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

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

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

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

Dividing $x^3 - 25x^2 + 67x + 3$ by $x^2 - 22x + 1$: quotient $x - 3$, remainder $6$

$x^3 - 25x^2 + 67x + 3 = (x^2 - 22x + 1)(x - 3) + 6$

$= 0 \times (x - 3) + 6 = 6$

Answer: $6$

(i) $\sqrt{12 + 2\sqrt{35}} = \sqrt{7 + 5 + 2\sqrt{7 \times 5}}$

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

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

(ii) $4\sqrt{5} = 2 \times 2\sqrt{5} = 2\sqrt{20}$

$\sqrt{12 - 4\sqrt{5}} = \sqrt{10 + 2 - 2\sqrt{10 \times 2}}$

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

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

($\sqrt{10} > \sqrt{2}$, so (ii) is positive.)

Answer: (i) $\sqrt{7} + \sqrt{5}$ (ii) $\sqrt{10} - \sqrt{2}$

$\displaystyle \sqrt{3 + \sqrt{5}} = \sqrt{\frac{2(3 + \sqrt{5})}{2}} = \frac{\sqrt{6 + 2\sqrt{5}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{5 + 1 + 2\sqrt{5 \times 1}}}{\sqrt{2}}$

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

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

Answer: $\displaystyle \frac{\sqrt{5} + 1}{\sqrt{2}}$