Rationalization and Surds: MCQ test, Set 2

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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.

$\sqrt[4]{81} =$
A $3$ B $\displaystyle \frac{81}{4}$ C $9$ D $12$

Answer and solution

Answer: (A) $3$

$\sqrt[4]{81} = \sqrt[4]{3 \times 3 \times 3 \times 3} = 3$
Which of the following is irrational but not a surd?
A $\sqrt{13}$ B $\sqrt[3]{49}$ C $\sqrt[5]{18}$ D $\pi$

Answer and solution

Answer: (D) $\pi$

$\pi$ is irrational, but it is not the root of a rational number.

The other three are surds.
Pick the mixed surd from the following.
A $3\sqrt[3]{4}$ B $6\sqrt{324}$ C $\sqrt[4]{81}$ D $\sqrt{8}$

Answer and solution

Answer: (A) $3\sqrt[3]{4}$

$3\sqrt[3]{4} = 3 \cdot \sqrt[3]{4}$ : a rational number times a surd.

$\sqrt{8}$ is a pure surd; $6\sqrt{324} = 108$ and $\sqrt[4]{81} = 3$ are rational.
The conjugate surd of $5 - \sqrt{13}$ is
A $-5 + \sqrt{13}$ B $-5 - \sqrt{13}$ C $\displaystyle \frac{1}{5 - \sqrt{13}}$ D $5 + \sqrt{13}$

Answer and solution

Answer: (D) $5 + \sqrt{13}$

Only the sign of the surd part changes: the conjugate of $5 - \sqrt{13}$ is $5 + \sqrt{13}$ .
$\sqrt{2} \times \sqrt{50} =$
A $2\sqrt{50}$ B $10$ C $\sqrt{52}$ D $100$

Answer and solution

Answer: (B) $10$

$\sqrt{2} \times \sqrt{50} = \sqrt{100} = \sqrt{10 \times 10} = 10$
$\sqrt{21} \times \sqrt{23} =$
A $21\sqrt{23}$ B $\sqrt{44}$ C $483$ D $\sqrt{483}$

Answer and solution

Answer: (D) $\sqrt{483}$

$\sqrt{21} \times \sqrt{23} = \sqrt{21 \times 23} = \sqrt{483}$ (multiply the numbers under the roots).
$\displaystyle \frac{\sqrt{23}}{\sqrt{15}} =$
A $\displaystyle \frac{23}{15}$ B $\sqrt{8}$ C $\displaystyle \sqrt{\frac{15}{23}}$ D $\displaystyle \sqrt{\frac{23}{15}}$

Answer and solution

Answer: (D) $\displaystyle \sqrt{\frac{23}{15}}$

$\displaystyle \frac{\sqrt{23}}{\sqrt{15}} = \sqrt{\frac{23}{15}}$ (divide the numbers under the roots)
$\sqrt[5]{9}$ in exponent form is
A $\displaystyle \frac{9}{5}$ B $\displaystyle 9^{\frac{1}{5}}$ C $9^{5}$ D $\displaystyle 5^{\frac{1}{9}}$

Answer and solution

Answer: (B) $\displaystyle 9^{\frac{1}{5}}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$ , so $\displaystyle \sqrt[5]{9} = 9^{\frac{1}{5}}$
$\sqrt[4]{8} \times \sqrt[4]{2} =$
A $2$ B $16$ C $\sqrt[4]{10}$ D $\sqrt[8]{16}$

Answer and solution

Answer: (A) $2$

$\sqrt[4]{8} \times \sqrt[4]{2} = \sqrt[4]{8 \times 2}$

$= \sqrt[4]{16} = \sqrt[4]{2^4} = 2$
If $(k\sqrt{2})^2 = 128$ and $k > 0$ , then $k =$
A $64$ B $8\sqrt{2}$ C $16$ D $8$

Answer and solution

Answer: (D) $8$

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

$2k^2 = 128$ , so $k^2 = 64$ and $k = 8$
$(4 + \sqrt{3})(5 + \sqrt{3}) =$
A $17 + 9\sqrt{3}$ B $23$ C $23 + 9\sqrt{3}$ D $23 - 9\sqrt{3}$

Answer and solution

Answer: (C) $23 + 9\sqrt{3}$

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

$= 23 + 9\sqrt{3}$
A rationalising factor of $20 + \sqrt{5}$ is
A $20 + \sqrt{5}$ B $\sqrt{5}$ C $20 - \sqrt{5}$ D $-\sqrt{5} - 20$

Answer and solution

Answer: (C) $20 - \sqrt{5}$

$(20 + \sqrt{5})(20 - \sqrt{5}) = 395$ , a rational number.

None of the other products is rational.
The rationalising factor of $\sqrt[3]{4} - \sqrt[3]{2}$ is
A $\sqrt[3]{4} + \sqrt[3]{2}$ B $(\sqrt[3]{4})^2 - \sqrt[3]{4} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^2$ C $(\sqrt[3]{4})^2 + (\sqrt[3]{2})^2$ D $(\sqrt[3]{4})^2 + \sqrt[3]{4} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^2$

Answer and solution

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

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

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

Product $= (\sqrt[3]{4})^3 - (\sqrt[3]{2})^3 = 4 - 2 = 2$
$\sqrt{20} + \sqrt{245} =$
A $\sqrt{265}$ B $9\sqrt{5}$ C $9\sqrt{10}$ D $14\sqrt{5}$

Answer and solution

Answer: (B) $9\sqrt{5}$

$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

$\sqrt{245} = \sqrt{49 \times 5} = 7\sqrt{5}$

$2\sqrt{5} + 7\sqrt{5} = 9\sqrt{5}$ (add like surds)
$(\sqrt{11})^3 =$
A $3\sqrt{11}$ B $121$ C $11\sqrt{11}$ D $11$

Answer and solution

Answer: (C) $11\sqrt{11}$

$(\sqrt{11})^3 = (\sqrt{11})^2 \times \sqrt{11} = 11\sqrt{11}$
$(8 - \sqrt{5})^2 =$
A $69 - 8\sqrt{5}$ B $69 - 16\sqrt{5}$ C $69$ D $89 - 16\sqrt{5}$

Answer and solution

Answer: (B) $69 - 16\sqrt{5}$

$(8 - \sqrt{5})^2 = 8^2 + (\sqrt{5})^2 - 2(8)(\sqrt{5}) = 69 - 16\sqrt{5}$
$(\sqrt{6} + \sqrt{3})^2 =$
A $45 + 6\sqrt{2}$ B $9 - 6\sqrt{2}$ C $9 + 6\sqrt{2}$ D $9$

Answer and solution

Answer: (C) $9 + 6\sqrt{2}$

$(\sqrt{6} + \sqrt{3})^2 = 6 + 3 + 2\sqrt{18} = 9 + 6\sqrt{2}$
$(10 + 2\sqrt{6})(10 - 2\sqrt{6}) =$
A $-44$ B $124$ C $76$ D $88$

Answer and solution

Answer: (C) $76$

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

$= 10^2 - (2\sqrt{6})^2 = 100 - 24 = 76$
$\displaystyle \frac{1}{5 - \sqrt{5}} =$
A $\displaystyle \frac{5 + \sqrt{5}}{30}$ B $\displaystyle -\frac{5 + \sqrt{5}}{20}$ C $\displaystyle \frac{5 - \sqrt{5}}{20}$ D $\displaystyle \frac{5 + \sqrt{5}}{20}$

Answer and solution

Answer: (D) $\displaystyle \frac{5 + \sqrt{5}}{20}$

Multiply above and below by $5 + \sqrt{5}$ : $(5 - \sqrt{5})(5 + \sqrt{5}) = 25 - 5 = 20$

$\displaystyle = \frac{5 + \sqrt{5}}{20}$
If $\displaystyle \frac{10 + 3\sqrt{5}}{4 - \sqrt{5}} = a + b\sqrt{5}$ , then $b =$
A $2$ B $3$ C $5$ D $-2$

Answer and solution

Answer: (A) $2$

$\displaystyle \frac{10 + 3\sqrt{5}}{4 - \sqrt{5}} = \frac{(10 + 3\sqrt{5})(4 + \sqrt{5})}{(4 - \sqrt{5})(4 + \sqrt{5})}$

Numerator: $40 + 10\sqrt{5} + 12\sqrt{5} + 15 = 55 + 22\sqrt{5}$

Denominator: $4^2 - (\sqrt{5})^2 = 16 - 5 = 11$

$\displaystyle \frac{55 + 22\sqrt{5}}{11} = 5 + 2\sqrt{5}$

So $a = 5$ , $b = 2$
$\displaystyle \frac{\sqrt{7} + \sqrt{2}}{\sqrt{7} - \sqrt{2}} + \frac{\sqrt{7} - \sqrt{2}}{\sqrt{7} + \sqrt{2}} =$
A $\displaystyle -\frac{18}{5}$ B $\displaystyle \frac{18}{5}$ C $\displaystyle \frac{9}{5}$ D $\displaystyle \frac{4\sqrt{14}}{5}$

Answer and solution

Answer: (B) $\displaystyle \frac{18}{5}$

$\displaystyle = \frac{(\sqrt{7} + \sqrt{2})^2 + (\sqrt{7} - \sqrt{2})^2}{(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2})} = \frac{18}{5}$

The $\sqrt{14}$ terms cancel and $(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2}) = 7 - 2 = 5$ .
$\displaystyle \frac{1}{\sqrt{17} + \sqrt{13}} + \frac{1}{\sqrt{17} - \sqrt{13}} =$
A $\displaystyle \frac{\sqrt{17}}{2}$ B $0$ C $\displaystyle \frac{\sqrt{13}}{2}$ D $\displaystyle \frac{\sqrt{17}}{15}$

Answer and solution

Answer: (A) $\displaystyle \frac{\sqrt{17}}{2}$

$\displaystyle = \frac{(\sqrt{17} - \sqrt{13}) + (\sqrt{17} + \sqrt{13})}{(\sqrt{17} + \sqrt{13})(\sqrt{17} - \sqrt{13})} = \frac{2\sqrt{17}}{4} = \frac{\sqrt{17}}{2}$
If $\displaystyle x = \frac{1}{15 + 4\sqrt{14}}$ , then $\displaystyle x + \frac{1}{x} =$
A $30$ B $8\sqrt{14}$ C $15$ D $450$

Answer and solution

Answer: (A) $30$

$x = 15 - 4\sqrt{14}$ and $\displaystyle \frac{1}{x} = 15 + 4\sqrt{14}$

$\displaystyle x + \frac{1}{x} = 30$
$\sqrt{10 + \sqrt{96}} =$
A $2 - \sqrt{6}$ B $\sqrt{6} + 2$ C $\sqrt{6} - 2$ D $\sqrt{24} + 1$

Answer and solution

Answer: (B) $\sqrt{6} + 2$

$\sqrt{96} = \sqrt{4 \times 24} = 2\sqrt{24}$

$\sqrt{10 + \sqrt{96}} = \sqrt{6 + 4 + 2\sqrt{6 \times 4}}$

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

$= \sqrt{(\sqrt{6} + 2)^2} = \sqrt{6} + 2$
$\sqrt{4 + \sqrt{15}} =$
A $\displaystyle \frac{\sqrt{5} + \sqrt{3}}{2}$ B $\displaystyle \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}}$ C $\displaystyle \frac{\sqrt{5} + \sqrt{3}}{\sqrt{2}}$ D $\sqrt{5} + \sqrt{3}$

Answer and solution

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

$\displaystyle \sqrt{4 + \sqrt{15}} = \sqrt{\frac{2(4 + \sqrt{15})}{2}} = \frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{5 + 3 + 2\sqrt{5 \times 3}}}{\sqrt{2}}$

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

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

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MCQ test, Set 2