Rationalization and Surds: MCQ test, Set 6

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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.

$\sqrt[4]{1296} =$
A $36$ B $6$ C $24$ D $324$

Answer and solution

Answer: (B) $6$

$\sqrt[4]{1296} = \sqrt[4]{6 \times 6 \times 6 \times 6} = 6$
$e = 2.71\ldots$ is
A a mixed surd B a rational number C an irrational number, but not a surd D a surd

Answer and solution

Answer: (C) 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.
$5\sqrt{n}$ is a mixed surd when $n$ is
A $225$ B $36$ C $17$ D $196$

Answer and solution

Answer: (C) $17$

$4^2 = 16$ and $5^2 = 25$ , so $17$ is not a perfect square. So $5\sqrt{17}$ is a mixed surd.

For the others $\sqrt{n}$ is a whole number, so the product is rational.
The conjugate surd of $\sqrt{3} + 14$ is
A $\sqrt{3} - 14$ B $-\sqrt{3} - 14$ C $\displaystyle \frac{1}{\sqrt{3} + 14}$ D $-\sqrt{3} + 14$

Answer and solution

Answer: (D) $-\sqrt{3} + 14$

Only the sign of the surd part changes: the conjugate of $\sqrt{3} + 14$ is $-\sqrt{3} + 14$ .
$\sqrt{40} \times \sqrt{90} =$
A $60$ B $3600$ C $\sqrt{130}$ D $40\sqrt{90}$

Answer and solution

Answer: (A) $60$

$\sqrt{40} \times \sqrt{90} = \sqrt{3600} = \sqrt{60 \times 60} = 60$
$\displaystyle \frac{\sqrt{14}}{\sqrt{5}} =$
A $\sqrt{9}$ B $\displaystyle \sqrt{\frac{5}{14}}$ C $\displaystyle \sqrt{\frac{14}{5}}$ D $\displaystyle \frac{14}{5}$

Answer and solution

Answer: (C) $\displaystyle \sqrt{\frac{14}{5}}$

$\displaystyle \frac{\sqrt{14}}{\sqrt{5}} = \sqrt{\frac{14}{5}}$ (divide the numbers under the roots)
$\displaystyle 81^{\frac{1}{4}} =$
A $9$ B $3$ C $\displaystyle \frac{81}{4}$ D $12$

Answer and solution

Answer: (B) $3$

$\displaystyle 81^{\frac{1}{4}} = \sqrt[4]{81} = \sqrt[4]{3^4} = 3$
$\sqrt[4]{12} \times \sqrt[4]{108} =$
A $1296$ B $6$ C $\sqrt[8]{1296}$ D $\sqrt[4]{120}$

Answer and solution

Answer: (B) $6$

$\sqrt[4]{12} \times \sqrt[4]{108} = \sqrt[4]{12 \times 108}$

$= \sqrt[4]{1296} = \sqrt[4]{6^4} = 6$
$(12\sqrt{19})^2 =$
A $144\sqrt{19}$ B $51984$ C $228$ D $2736$

Answer and solution

Answer: (D) $2736$

$(12\sqrt{19})^2 = 12^2 \times (\sqrt{19})^2$

$= 144 \times 19 = 2736$
$(8 + \sqrt{3})(7 + \sqrt{3}) =$
A $59 + 15\sqrt{3}$ B $59$ C $59 - 15\sqrt{3}$ D $53 + 15\sqrt{3}$

Answer and solution

Answer: (A) $59 + 15\sqrt{3}$

$(8 + \sqrt{3})(7 + \sqrt{3}) = 56 + 8\sqrt{3} + 7\sqrt{3} + 3$

$= 59 + 15\sqrt{3}$
A rationalising factor of $3 + \sqrt{30}$ is
A $\sqrt{30}$ B $-\sqrt{30} - 3$ C $3 + \sqrt{30}$ D $3 - \sqrt{30}$

Answer and solution

Answer: (D) $3 - \sqrt{30}$

$(3 + \sqrt{30})(3 - \sqrt{30}) = -21$ , a rational number.

None of the other products is rational.
The rationalising factor of $\sqrt[3]{17} - \sqrt[3]{9}$ is
A $(\sqrt[3]{17})^2 - \sqrt[3]{17} \cdot \sqrt[3]{9} + (\sqrt[3]{9})^2$ B $(\sqrt[3]{17})^2 + \sqrt[3]{17} \cdot \sqrt[3]{9} + (\sqrt[3]{9})^2$ C $\sqrt[3]{17} + \sqrt[3]{9}$ D $(\sqrt[3]{17})^2 + (\sqrt[3]{9})^2$

Answer and solution

Answer: (B) $(\sqrt[3]{17})^2 + \sqrt[3]{17} \cdot \sqrt[3]{9} + (\sqrt[3]{9})^2$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{17}$ , $y = \sqrt[3]{9}$ .

Rationalising factor: $(\sqrt[3]{17})^2 + \sqrt[3]{17} \cdot \sqrt[3]{9} + (\sqrt[3]{9})^2$

Product $= (\sqrt[3]{17})^3 - (\sqrt[3]{9})^3 = 17 - 9 = 8$
$\sqrt{192} + \sqrt{147} =$
A $15\sqrt{6}$ B $15\sqrt{3}$ C $56\sqrt{3}$ D $\sqrt{339}$

Answer and solution

Answer: (B) $15\sqrt{3}$

$\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$

$\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}$

$8\sqrt{3} + 7\sqrt{3} = 15\sqrt{3}$ (add like surds)
$\sqrt{176}$ in simplest form is
A $11\sqrt{16}$ B $16\sqrt{11}$ C $4\sqrt{11}$ D $4\sqrt{44}$

Answer and solution

Answer: (C) $4\sqrt{11}$

$\sqrt{176} = \sqrt{16 \times 11} = 4\sqrt{11}$
$(\sqrt{7})^{5} =$
A $343\sqrt{7}$ B $49$ C $49\sqrt{7}$ D $7\sqrt{7}$

Answer and solution

Answer: (C) $49\sqrt{7}$

$(\sqrt{7})^{5} = [(\sqrt{7})^2]^{2} \times \sqrt{7} = 7^{2} \times \sqrt{7} = 49\sqrt{7}$
$(\sqrt{10} + \sqrt{7})^2 =$
A $17 - 2\sqrt{70}$ B $149 + 2\sqrt{70}$ C $17$ D $17 + 2\sqrt{70}$

Answer and solution

Answer: (D) $17 + 2\sqrt{70}$

$(\sqrt{10} + \sqrt{7})^2 = 10 + 7 + 2\sqrt{70} = 17 + 2\sqrt{70}$
$(2 - \sqrt{15})^2 =$
A $19 - 4\sqrt{15}$ B $19 - 2\sqrt{15}$ C $229 - 4\sqrt{15}$ D $19$

Answer and solution

Answer: (A) $19 - 4\sqrt{15}$

$(2 - \sqrt{15})^2 = 2^2 + (\sqrt{15})^2 - 2(2)(\sqrt{15}) = 19 - 4\sqrt{15}$
$(\sqrt{17} - \sqrt{6})(\sqrt{17} + \sqrt{6}) =$
A $11$ B $253$ C $\sqrt{11}$ D $23$

Answer and solution

Answer: (A) $11$

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

$= (\sqrt{17})^2 - (\sqrt{6})^2 = 17 - 6 = 11$
Rationalising the denominator of $\displaystyle \frac{1}{\sqrt{19} - \sqrt{15}}$ gives
A $\displaystyle \frac{\sqrt{19} - \sqrt{15}}{4}$ B $\displaystyle \frac{\sqrt{19} + \sqrt{15}}{34}$ C $\displaystyle \frac{\sqrt{19} + \sqrt{15}}{4}$ D $\sqrt{19} + \sqrt{15}$

Answer and solution

Answer: (C) $\displaystyle \frac{\sqrt{19} + \sqrt{15}}{4}$

Multiply above and below by $\sqrt{19} + \sqrt{15}$ : $(\sqrt{19} - \sqrt{15})(\sqrt{19} + \sqrt{15}) = 19 - 15 = 4$

$\displaystyle = \frac{\sqrt{19} + \sqrt{15}}{4}$
If $\displaystyle \frac{6 + 2\sqrt{6}}{6 - 2\sqrt{6}} = a + b\sqrt{6}$ , then $a + b =$
A $10$ B $7$ C $3$ D $8$

Answer and solution

Answer: (B) $7$

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

Numerator: $36 + 12\sqrt{6} + 12\sqrt{6} + 24 = 60 + 24\sqrt{6}$

Denominator: $6^2 - (2\sqrt{6})^2 = 36 - 24 = 12$

$\displaystyle \frac{60 + 24\sqrt{6}}{12} = 5 + 2\sqrt{6}$

So $a = 5$ , $b = 2$
$\displaystyle \frac{9 + \sqrt{19}}{9 - \sqrt{19}} + \frac{9 - \sqrt{19}}{9 + \sqrt{19}} =$
A $\displaystyle \frac{18\sqrt{19}}{31}$ B $\displaystyle \frac{100}{31}$ C $\displaystyle \frac{50}{31}$ D $\displaystyle -\frac{100}{31}$

Answer and solution

Answer: (B) $\displaystyle \frac{100}{31}$

$\displaystyle = \frac{(9 + \sqrt{19})^2 + (9 - \sqrt{19})^2}{(9 + \sqrt{19})(9 - \sqrt{19})} = \frac{200}{62} = \frac{100}{31}$

The $\sqrt{19}$ terms cancel and $(9 + \sqrt{19})(9 - \sqrt{19}) = 62$ .
$\displaystyle \frac{1}{\sqrt{13} + \sqrt{6}} + \frac{1}{\sqrt{13} - \sqrt{6}} =$
A $\displaystyle \frac{2\sqrt{13}}{7}$ B $\displaystyle \frac{2\sqrt{6}}{7}$ C $\displaystyle \frac{2\sqrt{13}}{19}$ D $0$

Answer and solution

Answer: (A) $\displaystyle \frac{2\sqrt{13}}{7}$

$\displaystyle = \frac{(\sqrt{13} - \sqrt{6}) + (\sqrt{13} + \sqrt{6})}{(\sqrt{13} + \sqrt{6})(\sqrt{13} - \sqrt{6})} = \frac{2\sqrt{13}}{7}$
If $x = 6 + \sqrt{35}$ , then $x^2 - 12x =$
A $24\sqrt{35} - 1$ B $0$ C $-1$ D $1$

Answer and solution

Answer: (C) $-1$

$x - 6 = \sqrt{35}$ , so $(x - 6)^2 = 35$

$x^2 - 12x + 36 = 35$ , so $x^2 - 12x = -1$
$\sqrt{15 - \sqrt{56}} =$
A $1 - \sqrt{14}$ B $\sqrt{7} - \sqrt{2}$ C $\sqrt{14} + 1$ D $\sqrt{14} - 1$

Answer and solution

Answer: (D) $\sqrt{14} - 1$

$\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$

$\sqrt{15 - \sqrt{56}} = \sqrt{14 + 1 - 2\sqrt{14 \times 1}}$

$= \sqrt{(\sqrt{14})^2 + (1)^2 - 2(\sqrt{14})(1)}$

$= \sqrt{(\sqrt{14} - 1)^2} = \sqrt{14} - 1$
$\sqrt{7 + \sqrt{13}} =$
A $\displaystyle \frac{\sqrt{13} - 1}{\sqrt{2}}$ B $\sqrt{13} + 1$ C $\displaystyle \frac{\sqrt{13} + 1}{2}$ D $\displaystyle \frac{\sqrt{13} + 1}{\sqrt{2}}$

Answer and solution

Answer: (D) $\displaystyle \frac{\sqrt{13} + 1}{\sqrt{2}}$

$\displaystyle \sqrt{7 + \sqrt{13}} = \sqrt{\frac{2(7 + \sqrt{13})}{2}} = \frac{\sqrt{14 + 2\sqrt{13}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{13 + 1 + 2\sqrt{13 \times 1}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{13})^2 + (1)^2 + 2(\sqrt{13})(1)}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{13} + 1)^2}}{\sqrt{2}} = \frac{\sqrt{13} + 1}{\sqrt{2}}$

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MCQ test, Set 6