Question paper, Set 9

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

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

Answer: No. $\sqrt{49} = 7$, a rational number.

A surd $\sqrt[n]{a}$ is not a rational number, so every surd is irrational.

Answer: True

(i) $4\sqrt{43} = 4 \cdot \sqrt{43}$: a rational number times a surd, so a mixed surd

(ii) $5\sqrt{23} = 5 \cdot \sqrt{23}$: a rational number times a surd, so a mixed surd

(iii) $\sqrt{41}$ is a pure surd (no rational factor in front)

(iv) $8\sqrt{289} = 8 \times 17 = 136$, rational, not a surd

Mixed surds: (i) and (ii)

Answer: (i) and (ii)

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

The conjugate of $19 + \sqrt{17}$ is $19 - \sqrt{17}$.

Answer: $19 - \sqrt{17}$

$\sqrt{33} \times \sqrt{33} = (\sqrt{33})^2$

$= \sqrt{33^2} = 33$

Answer: $33$

(i) $\displaystyle \frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}}$

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

(ii) $\displaystyle \frac{\sqrt{224}}{\sqrt{14}} = \sqrt{\frac{224}{14}}$

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

Answer: (i) $2$ (ii) $4$

$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}$

$\sqrt[4]{11} \times \sqrt[4]{10} = \sqrt[4]{11 \times 10} = \sqrt[4]{110}$

Answer: $\sqrt[4]{110}$

$(k\sqrt{3})^2 = k^2 \times (\sqrt{3})^2 = 3k^2$

$3k^2 = 243$, so $k^2 = 81$

$k = 9$

Answer: $k = 9$

Take the conjugate: $8 + 5\sqrt{6}$.

$(8 - 5\sqrt{6})(8 + 5\sqrt{6}) = 8^2 - (5\sqrt{6})^2$

$(5\sqrt{6})^2 = 25 \times 6 = 150$

$= 64 - 150 = -86$, which is rational.

Answer: $8 + 5\sqrt{6}$; product $= -86$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{6}$, $y = \sqrt[3]{4}$.

Rationalising factor: $(\sqrt[3]{6})^2 - \sqrt[3]{6} \cdot \sqrt[3]{4} + (\sqrt[3]{4})^2$

Check: the product is $10$, a rational number.

Answer: $(\sqrt[3]{6})^2 - \sqrt[3]{6} \cdot \sqrt[3]{4} + (\sqrt[3]{4})^2$

$\sqrt{343} = \sqrt{49 \times 7} = 7\sqrt{7}$

Answer: $7\sqrt{7}$

(i) $(\sqrt{5})^{5} = [(\sqrt{5})^2]^{2} \times \sqrt{5} = 5^{2} \times \sqrt{5} = 25\sqrt{5}$

(ii) $(6\sqrt{6})^{3} = 6^{3} \times (\sqrt{6})^{3} = 216 \times 6\sqrt{6} = 1296\sqrt{6}$

(iii) $(4\sqrt{2})^{4} = 4^{4} \times (\sqrt{2})^{4} = 256 \times 4 = 1024$

Answer: (i) $25\sqrt{5}$ (ii) $1296\sqrt{6}$ (iii) $1024$

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

$(4\sqrt{2})^2 = 16 \times 2 = 32$

$2(4\sqrt{2})(\sqrt{7}) = 8\sqrt{14}$

$= 32 + 7 + 8\sqrt{14} = 39 + 8\sqrt{14}$

Answer: $39 + 8\sqrt{14}$

$(5 + 4\sqrt{3})(4 + \sqrt{3}) = 20 + 5\sqrt{3} + 16\sqrt{3} + 12$

[$4\sqrt{3} \times \sqrt{3} = 4 \times 3 = 12$]

$= 32 + 21\sqrt{3}$

Answer: $32 + 21\sqrt{3}$

$\displaystyle \frac{1}{8 - \sqrt{13}} = \frac{8 + \sqrt{13}}{(8 - \sqrt{13})(8 + \sqrt{13})}$

$(8 - \sqrt{13})(8 + \sqrt{13}) = 8^2 - (\sqrt{13})^2 = 64 - 13 = 51$

$\displaystyle = \frac{8 + \sqrt{13}}{51}$

Answer: $\displaystyle \frac{8 + \sqrt{13}}{51}$

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

$\displaystyle = \frac{15 + 13 + 2\sqrt{195}}{15 - 13}$

$= 14 + \sqrt{195}$

Comparing with $a + b\sqrt{195}$: $a = 14$, $b = 1$

Answer: $a = 14$, $b = 1$

$\displaystyle \frac{10 + \sqrt{5}}{10 - \sqrt{5}} - \frac{10 - \sqrt{5}}{10 + \sqrt{5}} = \frac{(10 + \sqrt{5})^2 - (10 - \sqrt{5})^2}{(10 - \sqrt{5})(10 + \sqrt{5})}$

$(10 + \sqrt{5})^2 = 105 + 20\sqrt{5}$

$(10 - \sqrt{5})^2 = 105 - 20\sqrt{5}$

$(10 + \sqrt{5})(10 - \sqrt{5}) = 100 - 5 = 95$

$\displaystyle = \frac{(105 + 20\sqrt{5}) - (105 - 20\sqrt{5})}{95} = \frac{8\sqrt{5}}{19}$

Answer: $\displaystyle \frac{8\sqrt{5}}{19}$

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

$\displaystyle = \frac{2\sqrt{2}}{25 - 2}$

$\displaystyle = \frac{2\sqrt{2}}{23}$

Answer: $\displaystyle \frac{2\sqrt{2}}{23}$

$\displaystyle x = \frac{1}{13 - 2\sqrt{42}} = \frac{13 + 2\sqrt{42}}{(13 - 2\sqrt{42})(13 + 2\sqrt{42})}$

$\displaystyle = \frac{13 + 2\sqrt{42}}{169 - 168} = 13 + 2\sqrt{42}$

$x - 13 = 2\sqrt{42}$

$(x - 13)^2 = (2\sqrt{42})^2 \Rightarrow x^2 - 26x + 169 = 168$

$\Rightarrow x^2 - 26x + 1 = 0$

Dividing $x^3 - 23x^2 - 77x + 8$ by $x^2 - 26x + 1$: quotient $x + 3$, remainder $5$

$x^3 - 23x^2 - 77x + 8 = (x^2 - 26x + 1)(x + 3) + 5$

$= 0 \times (x + 3) + 5 = 5$

Answer: $5$

Square both sides: $14 + a\sqrt{33} = (\sqrt{11} + \sqrt{3})^2$

$(\sqrt{11} + \sqrt{3})^2 = 11 + 3 + 2\sqrt{33} = 14 + 2\sqrt{33}$

$2\sqrt{33}$ is already in simplest form

Comparing: $a\sqrt{33} = 2\sqrt{33}$, so $a = 2$

Answer: $a = 2$