Rationalization and Surds: MCQ test, Set 3

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

For which value of $n$ is $\sqrt{n}$ a surd?
A $169$ B $400$ C $25$ D $19$

Answer and solution

Answer: (D) $19$

$4^2 = 16$ and $5^2 = 25$ , so $19$ is not a perfect square.

The other three are perfect squares, so their roots are rational.
$\pi = 3.14\ldots$ is
A an irrational number, but not a surd B a rational number C a mixed surd D a surd

Answer and solution

Answer: (A) an irrational number, but not a surd

Its decimal never ends and never repeats, so it is irrational.

It is not a root of a rational number, so it is not a surd.
$3\sqrt{n}$ is a mixed surd when $n$ is
A $40$ B $324$ C $64$ D $289$

Answer and solution

Answer: (A) $40$

$6^2 = 36$ and $7^2 = 49$ , so $40$ is not a perfect square. So $3\sqrt{40}$ is a mixed surd.

For the others $\sqrt{n}$ is a whole number, so the product is rational.
The conjugate surd of $7 - \sqrt{15}$ is
A $\displaystyle \frac{1}{7 - \sqrt{15}}$ B $-7 + \sqrt{15}$ C $-7 - \sqrt{15}$ D $7 + \sqrt{15}$

Answer and solution

Answer: (D) $7 + \sqrt{15}$

Only the sign of the surd part changes: the conjugate of $7 - \sqrt{15}$ is $7 + \sqrt{15}$ .
$\sqrt{15} \times \sqrt{15} =$
A $15$ B $2\sqrt{15}$ C $225$ D $\sqrt{15}$

Answer and solution

Answer: (A) $15$

$\sqrt{15} \times \sqrt{15} = \sqrt{15^2} = 15$
$\displaystyle \frac{\sqrt{17}}{\sqrt{10}} =$
A $\sqrt{7}$ B $\displaystyle \sqrt{\frac{17}{10}}$ C $\displaystyle \frac{17}{10}$ D $\displaystyle \sqrt{\frac{10}{17}}$

Answer and solution

Answer: (B) $\displaystyle \sqrt{\frac{17}{10}}$

$\displaystyle \frac{\sqrt{17}}{\sqrt{10}} = \sqrt{\frac{17}{10}}$ (divide the numbers under the roots)
$\displaystyle 1024^{\frac{1}{5}} =$
A $4$ B $20$ C $\displaystyle \frac{1024}{5}$ D $16$

Answer and solution

Answer: (A) $4$

$\displaystyle 1024^{\frac{1}{5}} = \sqrt[5]{1024} = \sqrt[5]{4^5} = 4$
$\sqrt[4]{27} \times \sqrt[4]{3} =$
A $81$ B $\sqrt[4]{30}$ C $3$ D $\sqrt[8]{81}$

Answer and solution

Answer: (C) $3$

$\sqrt[4]{27} \times \sqrt[4]{3} = \sqrt[4]{27 \times 3}$

$= \sqrt[4]{81} = \sqrt[4]{3^4} = 3$
If $(k\sqrt{17})^2 = 272$ and $k > 0$ , then $k =$
A $16$ B $8$ C $4$ D $4\sqrt{17}$

Answer and solution

Answer: (C) $4$

$(k\sqrt{17})^2 = k^2 \times 17 = 17k^2$

$17k^2 = 272$ , so $k^2 = 16$ and $k = 4$
$(5 - \sqrt{5})(1 - \sqrt{5}) =$
A $-6\sqrt{5}$ B $10 - 6\sqrt{5}$ C $10$ D $10 + 6\sqrt{5}$

Answer and solution

Answer: (B) $10 - 6\sqrt{5}$

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

$= 10 - 6\sqrt{5}$
$(3 + \sqrt{3})(6 + \sqrt{3}) =$
A $15 + 9\sqrt{3}$ B $21 - 9\sqrt{3}$ C $21 + 9\sqrt{3}$ D $21$

Answer and solution

Answer: (C) $21 + 9\sqrt{3}$

$(3 + \sqrt{3})(6 + \sqrt{3}) = 18 + 3\sqrt{3} + 6\sqrt{3} + 3$

$= 21 + 9\sqrt{3}$
A rationalising factor of $12 + \sqrt{6}$ is
A $12 - \sqrt{6}$ B $-\sqrt{6} - 12$ C $\sqrt{6}$ D $12 + \sqrt{6}$

Answer and solution

Answer: (A) $12 - \sqrt{6}$

$(12 + \sqrt{6})(12 - \sqrt{6}) = 138$ , a rational number.

None of the other products is rational.
The rationalising factor of $\sqrt[3]{15} + \sqrt[3]{14}$ is
A $(\sqrt[3]{15})^2 + \sqrt[3]{15} \cdot \sqrt[3]{14} + (\sqrt[3]{14})^2$ B $\sqrt[3]{15} - \sqrt[3]{14}$ C $(\sqrt[3]{15})^2 - \sqrt[3]{15} \cdot \sqrt[3]{14} + (\sqrt[3]{14})^2$ D $(\sqrt[3]{15})^2 + (\sqrt[3]{14})^2$

Answer and solution

Answer: (C) $(\sqrt[3]{15})^2 - \sqrt[3]{15} \cdot \sqrt[3]{14} + (\sqrt[3]{14})^2$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{15}$ , $y = \sqrt[3]{14}$ .

Rationalising factor: $(\sqrt[3]{15})^2 - \sqrt[3]{15} \cdot \sqrt[3]{14} + (\sqrt[3]{14})^2$

Product $= (\sqrt[3]{15})^3 + (\sqrt[3]{14})^3 = 15 + 14 = 29$
$\sqrt{175}$ in simplest form is
A $7\sqrt{5}$ B $25\sqrt{7}$ C $5\sqrt{35}$ D $5\sqrt{7}$

Answer and solution

Answer: (D) $5\sqrt{7}$

$\sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7}$
$(3\sqrt{2})^{4} =$
A $36$ B $162$ C $324$ D $12$

Answer and solution

Answer: (C) $324$

$(3\sqrt{2})^{4} = 3^{4} \times (\sqrt{2})^{4} = 81 \times 4 = 324$
$(4 - \sqrt{6})^2 =$
A $22$ B $22 - 4\sqrt{6}$ C $52 - 8\sqrt{6}$ D $22 - 8\sqrt{6}$

Answer and solution

Answer: (D) $22 - 8\sqrt{6}$

$(4 - \sqrt{6})^2 = 4^2 + (\sqrt{6})^2 - 2(4)(\sqrt{6}) = 22 - 8\sqrt{6}$
$(7 + 5\sqrt{6})(7 - 5\sqrt{6}) =$
A $-851$ B $199$ C $-101$ D $19$

Answer and solution

Answer: (C) $-101$

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

$= 7^2 - (5\sqrt{6})^2 = 49 - 150 = -101$
$\displaystyle \frac{1}{\sqrt{5} - \sqrt{3}} =$
A $\sqrt{5} + \sqrt{3}$ B $\displaystyle \frac{\sqrt{5} + \sqrt{3}}{8}$ C $\displaystyle \frac{\sqrt{5} - \sqrt{3}}{2}$ D $\displaystyle \frac{\sqrt{5} + \sqrt{3}}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{\sqrt{5} + \sqrt{3}}{2}$

Multiply above and below by $\sqrt{5} + \sqrt{3}$ : $(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = 5 - 3 = 2$

$\displaystyle = \frac{\sqrt{5} + \sqrt{3}}{2}$
If $\displaystyle \frac{12 + 7\sqrt{6}}{4 - \sqrt{6}} = a + b\sqrt{6}$ , then $a + b =$
A $19$ B $36$ C $5$ D $13$

Answer and solution

Answer: (D) $13$

$\displaystyle \frac{12 + 7\sqrt{6}}{4 - \sqrt{6}} = \frac{(12 + 7\sqrt{6})(4 + \sqrt{6})}{(4 - \sqrt{6})(4 + \sqrt{6})}$

Numerator: $48 + 12\sqrt{6} + 28\sqrt{6} + 42 = 90 + 40\sqrt{6}$

Denominator: $4^2 - (\sqrt{6})^2 = 16 - 6 = 10$

$\displaystyle \frac{90 + 40\sqrt{6}}{10} = 9 + 4\sqrt{6}$

So $a = 9$ , $b = 4$
$\displaystyle \frac{4 + \sqrt{7}}{4 - \sqrt{7}} + \frac{4 - \sqrt{7}}{4 + \sqrt{7}} =$
A $\displaystyle \frac{16\sqrt{7}}{9}$ B $\displaystyle \frac{46}{9}$ C $\displaystyle \frac{23}{9}$ D $\displaystyle -\frac{46}{9}$

Answer and solution

Answer: (B) $\displaystyle \frac{46}{9}$

$\displaystyle = \frac{(4 + \sqrt{7})^2 + (4 - \sqrt{7})^2}{(4 + \sqrt{7})(4 - \sqrt{7})} = \frac{46}{9}$

The $\sqrt{7}$ terms cancel and $(4 + \sqrt{7})(4 - \sqrt{7}) = 9$ .
$\displaystyle \frac{1}{\sqrt{6} + \sqrt{2}} + \frac{1}{\sqrt{6} - \sqrt{2}} =$
A $0$ B $\displaystyle \frac{\sqrt{2}}{2}$ C $\displaystyle \frac{\sqrt{6}}{4}$ D $\displaystyle \frac{\sqrt{6}}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{\sqrt{6}}{2}$

$\displaystyle = \frac{(\sqrt{6} - \sqrt{2}) + (\sqrt{6} + \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}$
$\displaystyle \frac{1}{3 + \sqrt{7}} + \frac{1}{3 - \sqrt{7}} =$
A $0$ B $\sqrt{7}$ C $3$ D $\displaystyle \frac{3}{8}$

Answer and solution

Answer: (C) $3$

$\displaystyle = \frac{(3 - \sqrt{7}) + (3 + \sqrt{7})}{(3 + \sqrt{7})(3 - \sqrt{7})} = \frac{6}{2} = 3$
If $x = \sqrt[3]{8 + 3\sqrt{7}}$ , then $\displaystyle x^3 + \frac{1}{x^3} =$
A $8$ B $6\sqrt{7}$ C $16$ D $0$

Answer and solution

Answer: (C) $16$

$x^3 = 8 + 3\sqrt{7}$ , so $\displaystyle \frac{1}{x^3} = \frac{1}{8 + 3\sqrt{7}} = 8 - 3\sqrt{7}$

$\displaystyle x^3 + \frac{1}{x^3} = (8 + 3\sqrt{7}) + (8 - 3\sqrt{7}) = 16$
$\sqrt{14 + 2\sqrt{33}} =$
A $\sqrt{3} - \sqrt{11}$ B $\sqrt{11} + \sqrt{3}$ C $\sqrt{11} - \sqrt{3}$ D $\sqrt{33} + 1$

Answer and solution

Answer: (B) $\sqrt{11} + \sqrt{3}$

$\sqrt{14 + 2\sqrt{33}} = \sqrt{11 + 3 + 2\sqrt{11 \times 3}}$

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

$= \sqrt{(\sqrt{11} + \sqrt{3})^2} = \sqrt{11} + \sqrt{3}$
$\sqrt{7 - \sqrt{45}} =$
A $\displaystyle \frac{3 + \sqrt{5}}{\sqrt{2}}$ B $\displaystyle \frac{3 - \sqrt{5}}{\sqrt{2}}$ C $3 - \sqrt{5}$ D $\displaystyle \frac{3 - \sqrt{5}}{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{3 - \sqrt{5}}{\sqrt{2}}$

$\displaystyle \sqrt{7 - \sqrt{45}} = \sqrt{\frac{2(7 - \sqrt{45})}{2}} = \frac{\sqrt{14 - 2\sqrt{45}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{9 + 5 - 2\sqrt{9 \times 5}}}{\sqrt{2}}$

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

$\displaystyle = \frac{\sqrt{(3 - \sqrt{5})^2}}{\sqrt{2}} = \frac{3 - \sqrt{5}}{\sqrt{2}}$

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MCQ test, Set 3