MCQ test, Set 5

Answer: (B) $\sqrt[5]{33}$

$2^5 = 32$ and $3^5 = 243$, so $33$ is not a perfect fifth power. So $\sqrt[5]{33}$ is a surd.

The others are rational: $\sqrt{36} = \sqrt{6 \times 6} = 6$, $\sqrt[5]{1024} = \sqrt[5]{4 \times 4 \times 4 \times 4 \times 4} = 4$, $\sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = 10$

Answer: (D) $\sqrt[3]{38}$ only

$3^3 = 27$ and $4^3 = 64$, so $38$ is not a perfect cube. So $\sqrt[3]{38}$ is a surd.

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

Answer: (D) $4\sqrt{23}$

$4\sqrt{23} = 4 \cdot \sqrt{23}$: a rational number times a surd.

$\sqrt{37}$ is a pure surd; $3\sqrt{25} = 15$ and $\sqrt[4]{81} = 3$ are rational.

Answer: (C) $12 - \sqrt{19}$

Only the sign of the surd part changes: the conjugate of $12 + \sqrt{19}$ is $12 - \sqrt{19}$.

Answer: (A) $\sqrt{210}$

$\sqrt{10} \times \sqrt{21} = \sqrt{10 \times 21} = \sqrt{210}$ (multiply the numbers under the roots).

Answer: (C) $2$

$\displaystyle \frac{\sqrt{68}}{\sqrt{17}} = \sqrt{\frac{68}{17}}$

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

Answer: (B) $\sqrt[4]{19}$

$\displaystyle a^{\frac{1}{n}} = \sqrt[n]{a}$, so $\displaystyle 19^{\frac{1}{4}} = \sqrt[4]{19}$

Answer: (C) $3$

$\displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{2}} = \sqrt[4]{\frac{162}{2}}$

$= \sqrt[4]{81} = \sqrt[4]{3^4} = 3$

Answer: (B) $5$

$(k\sqrt{10})^2 = k^2 \times 10 = 10k^2$

$10k^2 = 250$, so $k^2 = 25$ and $k = 5$

Answer: (B) $27 - 10\sqrt{6}$

$(7 - \sqrt{6})(3 - \sqrt{6}) = 21 - 7\sqrt{6} - 3\sqrt{6} + 6$

$= 27 - 10\sqrt{6}$

Answer: (A) $\sqrt{29}$

$2\sqrt{29} \times \sqrt{29} = 2 \times 29 = 58$, a rational number.

Answer: (B) $24$

Use $(x + y)(x^2 - xy + y^2) = x^3 + y^3$ with $x = \sqrt[3]{14}$, $y = \sqrt[3]{10}$.

Product $= (\sqrt[3]{14})^3 + (\sqrt[3]{10})^3 = 14 + 10 = 24$

Answer: (B) $12\sqrt{3}$

$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$

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

$5\sqrt{3} + 7\sqrt{3} = 12\sqrt{3}$ (add like surds)

Answer: (D) $16\sqrt{2}$

$(\sqrt{2})^{9} = [(\sqrt{2})^2]^{4} \times \sqrt{2} = 2^{4} \times \sqrt{2} = 16\sqrt{2}$

Answer: (A) $128\sqrt{2}$

$(4\sqrt{2})^{3} = 4^{3} \times (\sqrt{2})^{3} = 64 \times 2\sqrt{2} = 128\sqrt{2}$

Answer: (D) $62 + 14\sqrt{13}$

$(7 + \sqrt{13})^2 = 7^2 + (\sqrt{13})^2 + 2(7)(\sqrt{13}) = 62 + 14\sqrt{13}$

Answer: (A) $2$

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

$= (\sqrt{15})^2 - (\sqrt{13})^2 = 15 - 13 = 2$

Answer: (A) $\displaystyle \frac{\sqrt{6} + \sqrt{2}}{4}$

Multiply above and below by $\sqrt{6} + \sqrt{2}$: $(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2}) = 6 - 2 = 4$

$\displaystyle = \frac{\sqrt{6} + \sqrt{2}}{4}$

Answer: (C) $2$

$\displaystyle \frac{1 + \sqrt{7}}{3 - \sqrt{7}} = \frac{(1 + \sqrt{7})(3 + \sqrt{7})}{(3 - \sqrt{7})(3 + \sqrt{7})}$

Numerator: $3 + \sqrt{7} + 3\sqrt{7} + 7 = 10 + 4\sqrt{7}$

Denominator: $3^2 - (\sqrt{7})^2 = 9 - 7 = 2$

$\displaystyle \frac{10 + 4\sqrt{7}}{2} = 5 + 2\sqrt{7}$

So $a = 5$, $b = 2$

Answer: (D) $2$

$\displaystyle \frac{2 + 2\sqrt{7}}{6 - 2\sqrt{7}} = \frac{(2 + 2\sqrt{7})(6 + 2\sqrt{7})}{(6 - 2\sqrt{7})(6 + 2\sqrt{7})}$

Numerator: $12 + 4\sqrt{7} + 12\sqrt{7} + 28 = 40 + 16\sqrt{7}$

Denominator: $6^2 - (2\sqrt{7})^2 = 36 - 28 = 8$

$\displaystyle \frac{40 + 16\sqrt{7}}{8} = 5 + 2\sqrt{7}$

So $a = 5$, $b = 2$

Answer: (C) $\displaystyle \frac{46}{3}$

$\displaystyle = \frac{(\sqrt{13} + \sqrt{10})^2 + (\sqrt{13} - \sqrt{10})^2}{(\sqrt{13} + \sqrt{10})(\sqrt{13} - \sqrt{10})} = \frac{46}{3}$

The $\sqrt{130}$ terms cancel and $(\sqrt{13} + \sqrt{10})(\sqrt{13} - \sqrt{10}) = 13 - 10 = 3$.

Answer: (D) $\displaystyle \frac{\sqrt{19}}{8}$

$\displaystyle = \frac{(\sqrt{19} - \sqrt{3}) + (\sqrt{19} + \sqrt{3})}{(\sqrt{19} + \sqrt{3})(\sqrt{19} - \sqrt{3})} = \frac{2\sqrt{19}}{16} = \frac{\sqrt{19}}{8}$

Answer: (A) $-1$

$x - 5 = -2\sqrt{6}$, so $(x - 5)^2 = 24$

$x^2 - 10x + 25 = 24$, so $x^2 - 10x = -1$

Answer: (B) $\sqrt{11} - \sqrt{6}$

$\sqrt{17 - 2\sqrt{66}} = \sqrt{11 + 6 - 2\sqrt{11 \times 6}}$

$= \sqrt{(\sqrt{11})^2 + (\sqrt{6})^2 - 2(\sqrt{11})(\sqrt{6})}$

$= \sqrt{(\sqrt{11} - \sqrt{6})^2} = \sqrt{11} - \sqrt{6}$

Answer: (D) $\displaystyle \frac{\sqrt{7} - \sqrt{5}}{\sqrt{2}}$

$\displaystyle \sqrt{6 - \sqrt{35}} = \sqrt{\frac{2(6 - \sqrt{35})}{2}} = \frac{\sqrt{12 - 2\sqrt{35}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{7 + 5 - 2\sqrt{7 \times 5}}}{\sqrt{2}}$

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

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