Rationalization and Surds: MCQ test, Set 9

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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.

$\sqrt[5]{1024} =$
A $20$ B $\displaystyle \frac{1024}{5}$ C $16$ D $4$

Answer and solution

Answer: (D) $4$

$\sqrt[5]{1024} = \sqrt[5]{4 \times 4 \times 4 \times 4 \times 4} = 4$
$\sqrt[4]{25}$ and $\pi$ are both irrational. Which of them is a surd?
A Both B Neither C $\pi$ only D $\sqrt[4]{25}$ only

Answer and solution

Answer: (D) $\sqrt[4]{25}$ only

$2^4 = 16$ and $3^4 = 81$ , so $25$ is not a perfect fourth power. So $\sqrt[4]{25}$ is a surd.

$\pi$ is irrational but it is not a root of a rational number, so it is not a surd.
$7\sqrt[3]{n}$ is a mixed surd when $n$ is
A $512$ B $19$ C $1000$ D $216$

Answer and solution

Answer: (B) $19$

$2^3 = 8$ and $3^3 = 27$ , so $19$ is not a perfect cube. So $7\sqrt[3]{19}$ is a mixed surd.

For the others $\sqrt[3]{n}$ is a whole number, so the product is rational.
The conjugate of $\sqrt{17} + 20$ is
A $\displaystyle \frac{1}{\sqrt{17} + 20}$ B $-\sqrt{17} + 20$ C $-\sqrt{17} - 20$ D $\sqrt{17} - 20$

Answer and solution

Answer: (B) $-\sqrt{17} + 20$

Only the sign of the surd part changes: the conjugate of $\sqrt{17} + 20$ is $-\sqrt{17} + 20$ .
$(\sqrt{32})^2 =$
A $\sqrt{32}$ B $1024$ C $2\sqrt{32}$ D $32$

Answer and solution

Answer: (D) $32$

$\sqrt{32} \times \sqrt{32} = \sqrt{32^2} = 32$
$\displaystyle \frac{\sqrt{84}}{\sqrt{21}} =$
A $2\sqrt{21}$ B $2$ C $\sqrt{63}$ D $4$

Answer and solution

Answer: (B) $2$

$\displaystyle \frac{\sqrt{84}}{\sqrt{21}} = \sqrt{\frac{84}{21}}$

$= \sqrt{4} = \sqrt{2 \times 2} = 2$
$\displaystyle 729^{\frac{1}{6}} =$
A $\displaystyle \frac{729}{6}$ B $18$ C $3$ D $9$

Answer and solution

Answer: (C) $3$

$\displaystyle 729^{\frac{1}{6}} = \sqrt[6]{729} = \sqrt[6]{3^6} = 3$
$\displaystyle \frac{\sqrt[3]{875}}{\sqrt[3]{7}} =$
A $125$ B $5$ C $\sqrt[3]{882}$ D $\sqrt[3]{868}$

Answer and solution

Answer: (B) $5$

$\displaystyle \frac{\sqrt[3]{875}}{\sqrt[3]{7}} = \sqrt[3]{\frac{875}{7}}$

$= \sqrt[3]{125} = \sqrt[3]{5^3} = 5$
If $(k\sqrt{21})^2 = 756$ and $k > 0$ , then $k =$
A $6$ B $12$ C $36$ D $6\sqrt{21}$

Answer and solution

Answer: (A) $6$

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

$21k^2 = 756$ , so $k^2 = 36$ and $k = 6$
$(4 - \sqrt{13})(1 - \sqrt{13}) =$
A $17 + 5\sqrt{13}$ B $17 - 5\sqrt{13}$ C $-9 - 5\sqrt{13}$ D $17$

Answer and solution

Answer: (B) $17 - 5\sqrt{13}$

$(4 - \sqrt{13})(1 - \sqrt{13}) = 4 - 4\sqrt{13} - \sqrt{13} + 13$

$= 17 - 5\sqrt{13}$
A rationalising factor of $15 + \sqrt{13}$ is
A $-\sqrt{13} - 15$ B $\sqrt{13}$ C $15 + \sqrt{13}$ D $15 - \sqrt{13}$

Answer and solution

Answer: (D) $15 - \sqrt{13}$

$(15 + \sqrt{13})(15 - \sqrt{13}) = 212$ , a rational number.

None of the other products is rational.
The rationalising factor of $\sqrt[3]{11} + \sqrt[3]{3}$ is
A $(\sqrt[3]{11})^2 - \sqrt[3]{11} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$ B $(\sqrt[3]{11})^2 + (\sqrt[3]{3})^2$ C $\sqrt[3]{11} - \sqrt[3]{3}$ D $(\sqrt[3]{11})^2 + \sqrt[3]{11} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$

Answer and solution

Answer: (A) $(\sqrt[3]{11})^2 - \sqrt[3]{11} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{11}$ , $y = \sqrt[3]{3}$ .

Rationalising factor: $(\sqrt[3]{11})^2 - \sqrt[3]{11} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2$

Product $= (\sqrt[3]{11})^3 + (\sqrt[3]{3})^3 = 11 + 3 = 14$
$\sqrt{375}$ in simplest form is
A $25\sqrt{15}$ B $5\sqrt{75}$ C $15\sqrt{5}$ D $5\sqrt{15}$

Answer and solution

Answer: (D) $5\sqrt{15}$

$\sqrt{375} = \sqrt{25 \times 15} = 5\sqrt{15}$
$(2\sqrt{3})^{3} =$
A $6\sqrt{3}$ B $8\sqrt{3}$ C $24\sqrt{3}$ D $24$

Answer and solution

Answer: (C) $24\sqrt{3}$

$(2\sqrt{3})^{3} = 2^{3} \times (\sqrt{3})^{3} = 8 \times 3\sqrt{3} = 24\sqrt{3}$
$(10 - \sqrt{10})^2 =$
A $200 - 20\sqrt{10}$ B $110 - 10\sqrt{10}$ C $110 - 20\sqrt{10}$ D $110$

Answer and solution

Answer: (C) $110 - 20\sqrt{10}$

$(10 - \sqrt{10})^2 = 10^2 + (\sqrt{10})^2 - 2(10)(\sqrt{10}) = 110 - 20\sqrt{10}$
$(\sqrt{13} - \sqrt{11})(\sqrt{13} + \sqrt{11}) =$
A $48$ B $2$ C $24$ D $\sqrt{2}$

Answer and solution

Answer: (B) $2$

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

$= (\sqrt{13})^2 - (\sqrt{11})^2 = 13 - 11 = 2$
$(2 + 5\sqrt{5})(2 - 5\sqrt{5}) =$
A $-121$ B $129$ C $-621$ D $-21$

Answer and solution

Answer: (A) $-121$

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

$= 2^2 - (5\sqrt{5})^2 = 4 - 125 = -121$
$\displaystyle \frac{1}{\sqrt{11} + \sqrt{10}} =$
A $\displaystyle \frac{\sqrt{11} - \sqrt{10}}{21}$ B $\sqrt{11} + \sqrt{10}$ C $\sqrt{10} - \sqrt{11}$ D $\sqrt{11} - \sqrt{10}$

Answer and solution

Answer: (D) $\sqrt{11} - \sqrt{10}$

Multiply above and below by $\sqrt{11} - \sqrt{10}$ : $(\sqrt{11} + \sqrt{10})(\sqrt{11} - \sqrt{10}) = 11 - 10 = 1$

$= \sqrt{11} - \sqrt{10}$
If $\displaystyle \frac{2 + 3\sqrt{2}}{2 - \sqrt{2}} = a + b\sqrt{2}$ , then $b =$
A $4$ B $5$ C $-4$ D $3$

Answer and solution

Answer: (A) $4$

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

Numerator: $4 + 2\sqrt{2} + 6\sqrt{2} + 6 = 10 + 8\sqrt{2}$

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

$\displaystyle \frac{10 + 8\sqrt{2}}{2} = 5 + 4\sqrt{2}$

So $a = 5$ , $b = 4$
$\displaystyle \frac{\sqrt{17} + \sqrt{15}}{\sqrt{17} - \sqrt{15}} + \frac{\sqrt{17} - \sqrt{15}}{\sqrt{17} + \sqrt{15}} =$
A $2\sqrt{255}$ B $-32$ C $16$ D $32$

Answer and solution

Answer: (D) $32$

$\displaystyle = \frac{(\sqrt{17} + \sqrt{15})^2 + (\sqrt{17} - \sqrt{15})^2}{(\sqrt{17} + \sqrt{15})(\sqrt{17} - \sqrt{15})} = \frac{64}{2} = 32$

The $\sqrt{255}$ terms cancel and $(\sqrt{17} + \sqrt{15})(\sqrt{17} - \sqrt{15}) = 17 - 15 = 2$ .
$\displaystyle \frac{12 + \sqrt{13}}{12 - \sqrt{13}} + \frac{12 - \sqrt{13}}{12 + \sqrt{13}} =$
A $\displaystyle \frac{314}{131}$ B $\displaystyle \frac{48\sqrt{13}}{131}$ C $\displaystyle -\frac{314}{131}$ D $\displaystyle \frac{157}{131}$

Answer and solution

Answer: (A) $\displaystyle \frac{314}{131}$

$\displaystyle = \frac{(12 + \sqrt{13})^2 + (12 - \sqrt{13})^2}{(12 + \sqrt{13})(12 - \sqrt{13})} = \frac{314}{131}$

The $\sqrt{13}$ terms cancel and $(12 + \sqrt{13})(12 - \sqrt{13}) = 131$ .
$\displaystyle \frac{1}{5 + \sqrt{3}} + \frac{1}{5 - \sqrt{3}} =$
A $0$ B $\displaystyle \frac{5}{14}$ C $\displaystyle \frac{5}{11}$ D $\displaystyle \frac{\sqrt{3}}{11}$

Answer and solution

Answer: (C) $\displaystyle \frac{5}{11}$

$\displaystyle = \frac{(5 - \sqrt{3}) + (5 + \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})} = \frac{10}{22} = \frac{5}{11}$
If $x = \sqrt[3]{10 + 3\sqrt{11}}$ , then $\displaystyle x^3 + \frac{1}{x^3} =$
A $6\sqrt{11}$ B $10$ C $20$ D $0$

Answer and solution

Answer: (C) $20$

$x^3 = 10 + 3\sqrt{11}$ , so $\displaystyle \frac{1}{x^3} = \frac{1}{10 + 3\sqrt{11}} = 10 - 3\sqrt{11}$

$\displaystyle x^3 + \frac{1}{x^3} = (10 + 3\sqrt{11}) + (10 - 3\sqrt{11}) = 20$
$\sqrt{16 + \sqrt{220}} =$
A $\sqrt{11} + \sqrt{5}$ B $\sqrt{11} - \sqrt{5}$ C $\sqrt{5} - \sqrt{11}$ D $\sqrt{55} + 1$

Answer and solution

Answer: (A) $\sqrt{11} + \sqrt{5}$

$\sqrt{220} = \sqrt{4 \times 55} = 2\sqrt{55}$

$\sqrt{16 + \sqrt{220}} = \sqrt{11 + 5 + 2\sqrt{11 \times 5}}$

$= \sqrt{(\sqrt{11})^2 + (\sqrt{5})^2 + 2(\sqrt{11})(\sqrt{5})}$

$= \sqrt{(\sqrt{11} + \sqrt{5})^2} = \sqrt{11} + \sqrt{5}$
$\sqrt{12 + \sqrt{23}} =$
A $\displaystyle \frac{\sqrt{23} + 1}{2}$ B $\displaystyle \frac{\sqrt{23} - 1}{\sqrt{2}}$ C $\sqrt{23} + 1$ D $\displaystyle \frac{\sqrt{23} + 1}{\sqrt{2}}$

Answer and solution

Answer: (D) $\displaystyle \frac{\sqrt{23} + 1}{\sqrt{2}}$

$\displaystyle \sqrt{12 + \sqrt{23}} = \sqrt{\frac{2(12 + \sqrt{23})}{2}} = \frac{\sqrt{24 + 2\sqrt{23}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{23 + 1 + 2\sqrt{23 \times 1}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23})^2 + (1)^2 + 2(\sqrt{23})(1)}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{23} + 1)^2}}{\sqrt{2}} = \frac{\sqrt{23} + 1}{\sqrt{2}}$

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MCQ test, Set 9