MCQ test, Set 7

Answer: (D) $\sqrt[5]{10}$

$1^5 = 1$ and $2^5 = 32$, so $10$ is not a perfect fifth power. So $\sqrt[5]{10}$ is a surd.

The others are rational: $\sqrt[5]{1024} = \sqrt[5]{4 \times 4 \times 4 \times 4 \times 4} = 4$, $\sqrt[5]{243} = \sqrt[5]{3 \times 3 \times 3 \times 3 \times 3} = 3$, $\sqrt[6]{64} = \sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2} = 2$

Answer: (D) Every surd is an irrational number.

A surd is never rational, so every surd is irrational.

$\pi$ is irrational but not a surd, so the second statement is false.

Answer: (B) $27$

$5^2 = 25$ and $6^2 = 36$, so $27$ is not a perfect square. So $4\sqrt{27}$ is a mixed surd.

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

Answer: (D) $9 - \sqrt{14}$

Only the sign of the surd part changes: the conjugate of $9 + \sqrt{14}$ is $9 - \sqrt{14}$.

Answer: (B) $27$

$\sqrt{27} \times \sqrt{27} = \sqrt{27^2} = 27$

Answer: (C) $\displaystyle \sqrt{\frac{19}{11}}$

$\displaystyle \frac{\sqrt{19}}{\sqrt{11}} = \sqrt{\frac{19}{11}}$ (divide the numbers under the roots)

Answer: (A) $5$

$\displaystyle 125^{\frac{1}{3}} = \sqrt[3]{125} = \sqrt[3]{5^3} = 5$

Answer: (C) $7$

$\sqrt[3]{7} \times \sqrt[3]{49} = \sqrt[3]{7 \times 49}$

$= \sqrt[3]{343} = \sqrt[3]{7^3} = 7$

Answer: (C) $3$

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

$14k^2 = 126$, so $k^2 = 9$ and $k = 3$

Answer: (D) $16 + 9\sqrt{2}$

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

$= 16 + 9\sqrt{2}$

Answer: (B) $18 - \sqrt{22}$

$(18 + \sqrt{22})(18 - \sqrt{22}) = 302$, a rational number.

None of the other products is rational.

Answer: (D) $(\sqrt[3]{9})^2 + \sqrt[3]{9} \cdot \sqrt[3]{7} + (\sqrt[3]{7})^2$

Use $(x - y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = \sqrt[3]{9}$, $y = \sqrt[3]{7}$.

Rationalising factor: $(\sqrt[3]{9})^2 + \sqrt[3]{9} \cdot \sqrt[3]{7} + (\sqrt[3]{7})^2$

Product $= (\sqrt[3]{9})^3 - (\sqrt[3]{7})^3 = 9 - 7 = 2$

Answer: (C) $12\sqrt{2}$

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

$\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$

$4\sqrt{2} + 8\sqrt{2} = 12\sqrt{2}$ (add like surds)

Answer: (B) $625\sqrt{5}$

$(5\sqrt{5})^{3} = 5^{3} \times (\sqrt{5})^{3} = 125 \times 5\sqrt{5} = 625\sqrt{5}$

Answer: (C) $11 - 2\sqrt{30}$

$(\sqrt{6} - \sqrt{5})^2 = 6 + 5 - 2\sqrt{30} = 11 - 2\sqrt{30}$

Answer: (D) $6$

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

$= 9^2 - (5\sqrt{3})^2 = 81 - 75 = 6$

Answer: (A) $\displaystyle \frac{6 + \sqrt{7}}{29}$

Multiply above and below by $6 + \sqrt{7}$: $(6 - \sqrt{7})(6 + \sqrt{7}) = 36 - 7 = 29$

$\displaystyle = \frac{6 + \sqrt{7}}{29}$

Answer: (C) $\displaystyle \frac{\sqrt{15} - \sqrt{3}}{12}$

Multiply above and below by $\sqrt{15} - \sqrt{3}$: $(\sqrt{15} + \sqrt{3})(\sqrt{15} - \sqrt{3}) = 15 - 3 = 12$

$\displaystyle = \frac{\sqrt{15} - \sqrt{3}}{12}$

Answer: (D) $3$

$\displaystyle \frac{6 + 5\sqrt{2}}{3 - \sqrt{2}} = \frac{(6 + 5\sqrt{2})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}$

Numerator: $18 + 6\sqrt{2} + 15\sqrt{2} + 10 = 28 + 21\sqrt{2}$

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

$\displaystyle \frac{28 + 21\sqrt{2}}{7} = 4 + 3\sqrt{2}$

So $a = 4$, $b = 3$

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

$\displaystyle = \frac{(2 + \sqrt{14})^2 + (2 - \sqrt{14})^2}{(2 + \sqrt{14})(2 - \sqrt{14})} = \frac{36}{-10} = -\frac{18}{5}$

The $\sqrt{14}$ terms cancel and $(2 + \sqrt{14})(2 - \sqrt{14}) = -10$.

Answer: (A) $\displaystyle \frac{8}{3}$

$\displaystyle = \frac{(4 - \sqrt{13}) + (4 + \sqrt{13})}{(4 + \sqrt{13})(4 - \sqrt{13})} = \frac{8}{3}$

Answer: (B) $16$

$x^3 = 8 - 3\sqrt{7}$, so $\displaystyle \frac{1}{x^3} = \frac{1}{8 - 3\sqrt{7}} = 8 + 3\sqrt{7}$

$\displaystyle x^3 + \frac{1}{x^3} = (8 - 3\sqrt{7}) + (8 + 3\sqrt{7}) = 16$

Answer: (B) $-1$

$x - 3 = 2\sqrt{2}$, so $(x - 3)^2 = 8$

$x^2 - 6x + 9 = 8$, so $x^2 - 6x = -1$

Answer: (A) $\sqrt{7} - \sqrt{6}$

$\sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}$

$\sqrt{13 - \sqrt{168}} = \sqrt{7 + 6 - 2\sqrt{7 \times 6}}$

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

$= \sqrt{(\sqrt{7} - \sqrt{6})^2} = \sqrt{7} - \sqrt{6}$

Answer: (B) $\displaystyle \frac{\sqrt{19} + 3}{\sqrt{2}}$

$\displaystyle \sqrt{14 + \sqrt{171}} = \sqrt{\frac{2(14 + \sqrt{171})}{2}} = \frac{\sqrt{28 + 2\sqrt{171}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{19 + 9 + 2\sqrt{19 \times 9}}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{19})^2 + (3)^2 + 2(\sqrt{19})(3)}}{\sqrt{2}}$

$\displaystyle = \frac{\sqrt{(\sqrt{19} + 3)^2}}{\sqrt{2}} = \frac{\sqrt{19} + 3}{\sqrt{2}}$