MCQ test, Set 8

Answer: (B) $400$

$\sqrt{400} = \sqrt{20 \times 20} = 20$

The other three are not perfect squares, so their roots are surds.

Answer: (B) $\sqrt[5]{19}$ only

$1^5 = 1$ and $2^5 = 32$, so $19$ is not a perfect fifth power. So $\sqrt[5]{19}$ is a surd.

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

Answer: (A) $24$

$4^2 = 16$ and $5^2 = 25$, so $24$ is not a perfect square. So $2\sqrt{24}$ is a mixed surd.

For the others $\sqrt{n}$ is a whole number, so the product is rational.

Answer: (D) $20 - \sqrt{21}$

Only the sign of the surd part changes: the conjugate of $20 + \sqrt{21}$ is $20 - \sqrt{21}$.

Answer: (C) $\sqrt{119}$

$\sqrt{7} \times \sqrt{17} = \sqrt{7 \times 17} = \sqrt{119}$ (multiply the numbers under the roots).

Answer: (C) $6$

$\displaystyle \frac{\sqrt{180}}{\sqrt{5}} = \sqrt{\frac{180}{5}}$

$= \sqrt{36} = \sqrt{6 \times 6} = 6$

Answer: (B) $\displaystyle \sqrt{\frac{13}{23}}$

$\displaystyle \frac{\sqrt{13}}{\sqrt{23}} = \sqrt{\frac{13}{23}}$ (divide the numbers under the roots)

Answer: (D) $\displaystyle 16^{\frac{1}{7}}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$, so $\displaystyle \sqrt[7]{16} = 16^{\frac{1}{7}}$

Answer: (B) $4$

$\displaystyle \frac{\sqrt[4]{768}}{\sqrt[4]{3}} = \sqrt[4]{\frac{768}{3}}$

$= \sqrt[4]{256} = \sqrt[4]{4^4} = 4$

Answer: (A) $686$

$(7\sqrt{14})^2 = 7^2 \times (\sqrt{14})^2$

$= 49 \times 14 = 686$

Answer: (D) $65 - 15\sqrt{11}$

$(6 - \sqrt{11})(9 - \sqrt{11}) = 54 - 6\sqrt{11} - 9\sqrt{11} + 11$

$= 65 - 15\sqrt{11}$

Answer: (C) $69 - 16\sqrt{6}$

$(9 - \sqrt{6})(7 - \sqrt{6}) = 63 - 9\sqrt{6} - 7\sqrt{6} + 6$

$= 69 - 16\sqrt{6}$

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

$4\sqrt{11} \times \sqrt{11} = 4 \times 11 = 44$, a rational number.

Answer: (C) $2$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{7}$, $y = \sqrt[3]{5}$.

Product $= (\sqrt[3]{7})^3 - (\sqrt[3]{5})^3 = 7 - 5 = 2$

Answer: (A) $4\sqrt{14}$

$\sqrt{224} = \sqrt{16 \times 14} = 4\sqrt{14}$

Answer: (C) $6\sqrt{6}$

$(\sqrt{6})^3 = (\sqrt{6})^2 \times \sqrt{6} = 6\sqrt{6}$

Answer: (C) $39 - 12\sqrt{3}$

$(6 - \sqrt{3})^2 = 6^2 + (\sqrt{3})^2 - 2(6)(\sqrt{3}) = 39 - 12\sqrt{3}$

Answer: (A) $50$

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

$= 10^2 - (5\sqrt{2})^2 = 100 - 50 = 50$

Answer: (D) $\displaystyle \frac{9 + \sqrt{3}}{78}$

Multiply above and below by $9 + \sqrt{3}$: $(9 - \sqrt{3})(9 + \sqrt{3}) = 81 - 3 = 78$

$\displaystyle = \frac{9 + \sqrt{3}}{78}$

Answer: (D) $11$

$\displaystyle \frac{4 + 2\sqrt{3}}{4 - 2\sqrt{3}} = \frac{(4 + 2\sqrt{3})(4 + 2\sqrt{3})}{(4 - 2\sqrt{3})(4 + 2\sqrt{3})}$

Numerator: $16 + 8\sqrt{3} + 8\sqrt{3} + 12 = 28 + 16\sqrt{3}$

Denominator: $4^2 - (2\sqrt{3})^2 = 16 - 12 = 4$

$\displaystyle \frac{28 + 16\sqrt{3}}{4} = 7 + 4\sqrt{3}$

So $a = 7$, $b = 4$

Answer: (C) $\displaystyle \frac{34}{5}$

$\displaystyle = \frac{(\sqrt{11} + \sqrt{6})^2 + (\sqrt{11} - \sqrt{6})^2}{(\sqrt{11} + \sqrt{6})(\sqrt{11} - \sqrt{6})} = \frac{34}{5}$

The $\sqrt{66}$ terms cancel and $(\sqrt{11} + \sqrt{6})(\sqrt{11} - \sqrt{6}) = 11 - 6 = 5$.

Answer: (A) $\displaystyle \frac{12}{31}$

$\displaystyle = \frac{(6 - \sqrt{5}) + (6 + \sqrt{5})}{(6 + \sqrt{5})(6 - \sqrt{5})} = \frac{12}{31}$

Answer: (C) $26$

$x^3 = 13 - 2\sqrt{42}$, so $\displaystyle \frac{1}{x^3} = \frac{1}{13 - 2\sqrt{42}} = 13 + 2\sqrt{42}$

$\displaystyle x^3 + \frac{1}{x^3} = (13 - 2\sqrt{42}) + (13 + 2\sqrt{42}) = 26$

Answer: (D) $3 - \sqrt{5}$

$\sqrt{14 - 2\sqrt{45}} = \sqrt{9 + 5 - 2\sqrt{9 \times 5}}$

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

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

Answer: (B) $\displaystyle \frac{\sqrt{11} + 3}{\sqrt{2}}$

$\displaystyle \sqrt{10 + \sqrt{99}} = \sqrt{\frac{2(10 + \sqrt{99})}{2}} = \frac{\sqrt{20 + 2\sqrt{99}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{11 + 9 + 2\sqrt{11 \times 9}}}{\sqrt{2}}$

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

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