Question paper, Set 4

$2^4 = 16$ and $3^4 = 81$, so $26$ is not a perfect fourth power.

So $\sqrt[4]{26}$ is not a rational number; it is a surd.

Answer: Yes. $26$ is not a perfect fourth power, so $\sqrt[4]{26}$ is not rational.

$\pi = 3.14\ldots$ is a non-terminating, non-recurring decimal, so it is irrational.

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

Answer: False

(i) $2\sqrt{100} = 2 \times 10 = 20$, rational, not a surd

(ii) $6\sqrt[3]{60} = 6 \cdot \sqrt[3]{60}$: a rational number times a surd, so a mixed surd

(iii) $3\sqrt{22} = 3 \cdot \sqrt{22}$: a rational number times a surd, so a mixed surd

(iv) $\sqrt[4]{17}$ is a pure surd (no rational factor in front)

Mixed surds: (ii) and (iii)

Answer: (ii) and (iii)

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

The conjugate of $2 + \sqrt{10}$ is $2 - \sqrt{10}$.

Answer: $2 - \sqrt{10}$

$\sqrt{3} \times \sqrt{22} = \sqrt{3 \times 22}$

$= \sqrt{66}$

Answer: $\sqrt{66}$

(i) $\displaystyle \frac{\sqrt{24}}{\sqrt{6}} = \sqrt{\frac{24}{6}}$

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

(ii) $\displaystyle \frac{\sqrt{325}}{\sqrt{13}} = \sqrt{\frac{325}{13}}$

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

Answer: (i) $2$ (ii) $5$

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

$\sqrt[3]{6} \times \sqrt[3]{11} = \sqrt[3]{6 \times 11} = \sqrt[3]{66}$

Answer: $\sqrt[3]{66}$

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

$23k^2 = 92$, so $k^2 = 4$

$k = 2$

Answer: $k = 2$

Take the conjugate: $4 - 2\sqrt{7}$.

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

$(2\sqrt{7})^2 = 4 \times 7 = 28$

$= 16 - 28 = -12$, which is rational.

Answer: $4 - 2\sqrt{7}$; product $= -12$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{17}$, $y = \sqrt[3]{5}$.

Rationalising factor: $(\sqrt[3]{17})^2 - \sqrt[3]{17} \cdot \sqrt[3]{5} + (\sqrt[3]{5})^2$

Check: the product is $22$, a rational number.

Answer: $(\sqrt[3]{17})^2 - \sqrt[3]{17} \cdot \sqrt[3]{5} + (\sqrt[3]{5})^2$

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

Answer: $3\sqrt{6}$

(i) $(\sqrt{3})^{9} = [(\sqrt{3})^2]^{4} \times \sqrt{3} = 3^{4} \times \sqrt{3} = 81\sqrt{3}$

(ii) $(3\sqrt{5})^{3} = 3^{3} \times (\sqrt{5})^{3} = 27 \times 5\sqrt{5} = 135\sqrt{5}$

(iii) $(2\sqrt{10})^{4} = 2^{4} \times (\sqrt{10})^{4} = 16 \times 100 = 1600$

Answer: (i) $81\sqrt{3}$ (ii) $135\sqrt{5}$ (iii) $1600$

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

$(4\sqrt{5})^2 = 16 \times 5 = 80$

$2(4\sqrt{5})(\sqrt{2}) = 8\sqrt{10}$

$= 80 + 2 + 8\sqrt{10} = 82 + 8\sqrt{10}$

Answer: $82 + 8\sqrt{10}$

$(6 + 4\sqrt{5})(7 - \sqrt{5}) = 42 - 6\sqrt{5} + 28\sqrt{5} - 20$

[$4\sqrt{5} \times (-\sqrt{5}) = -4 \times 5 = -20$]

$= 22 + 22\sqrt{5}$

Answer: $22 + 22\sqrt{5}$

$\displaystyle \frac{1}{\sqrt{14} - \sqrt{7}} = \frac{\sqrt{14} + \sqrt{7}}{(\sqrt{14} - \sqrt{7})(\sqrt{14} + \sqrt{7})}$

$(\sqrt{14} - \sqrt{7})(\sqrt{14} + \sqrt{7}) = 14 - 7 = 7$

$\displaystyle = \frac{\sqrt{14} + \sqrt{7}}{7}$

Answer: $\displaystyle \frac{\sqrt{14} + \sqrt{7}}{7}$

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

$\displaystyle = \frac{3 + 2 + 2\sqrt{6}}{3 - 2}$

$= 5 + 2\sqrt{6}$

Comparing with $a + b\sqrt{6}$: $a = 5$, $b = 2$

Answer: $a = 5$, $b = 2$

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

$(2 + 4\sqrt{7})^2 = 116 + 16\sqrt{7}$

$(2 - 4\sqrt{7})^2 = 116 - 16\sqrt{7}$

$(2 + 4\sqrt{7})(2 - 4\sqrt{7}) = -108$

$\displaystyle = \frac{116 + 16\sqrt{7}}{-108} + \frac{116 - 16\sqrt{7}}{-108}$

$\displaystyle = \frac{116 + 16\sqrt{7} + 116 - 16\sqrt{7}}{-108} = -\frac{58}{27}$

Answer: $\displaystyle -\frac{58}{27}$

$\displaystyle \frac{1}{8 - \sqrt{5}} - \frac{1}{8 + \sqrt{5}} = \frac{(8 + \sqrt{5}) - (8 - \sqrt{5})}{(8 - \sqrt{5})(8 + \sqrt{5})}$

$\displaystyle = \frac{2\sqrt{5}}{64 - 5}$

$\displaystyle = \frac{2\sqrt{5}}{59}$

Answer: $\displaystyle \frac{2\sqrt{5}}{59}$

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

$\displaystyle = \frac{4 + \sqrt{15}}{16 - 15} = 4 + \sqrt{15}$

$x - 4 = \sqrt{15}$

$(x - 4)^2 = (\sqrt{15})^2 \Rightarrow x^2 - 8x + 16 = 15$

$\Rightarrow x^2 - 8x + 1 = 0$

Dividing $x^3 - 10x^2 + 17x + 1$ by $x^2 - 8x + 1$: quotient $x - 2$, remainder $3$

$x^3 - 10x^2 + 17x + 1 = (x^2 - 8x + 1)(x - 2) + 3$

$= 0 \times (x - 2) + 3 = 3$

Answer: $3$

Square both sides: $8 - a\sqrt{15} = (\sqrt{5} - \sqrt{3})^2$

$(\sqrt{5} - \sqrt{3})^2 = 5 + 3 - 2\sqrt{15} = 8 - 2\sqrt{15}$

$2\sqrt{15}$ is already in simplest form

Comparing: $a\sqrt{15} = 2\sqrt{15}$, so $a = 2$

Answer: $a = 2$