$\sqrt{6} > \sqrt[3]{7}$

True

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False

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Solution

The correct option is A
True

Given surds are not 'Like surds'. So, to compare them the powers have to be made equal.

Now LCM of 2 and 3 is 6.

$⟹$ $\sqrt{6}$ = $6^{\frac{1}{2} \times \frac{3}{3}}$

= $\sqrt[6]{6^{3}}$

= $\sqrt[6]{216}$ -------------(1)

$\sqrt[3]{7}$ = $7^{\frac{1}{3} \times \frac{2}{2}}$

= $\sqrt[6]{7^{2}}$

= $\sqrt[6]{49}$ -------------(2)

From (1) and (2), $\sqrt[6]{216}$ > $\sqrt[6]{49}$

$∴$ $\sqrt{6} > \sqrt[3]{7}$

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Similar questions

Question 6

Which of the following expressions show that rational numbers are associative under multiplication.
(a) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5}$
(b) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = \frac{2}{3} \times (\frac{3}{5} \times \frac{- 6}{7})$
(c) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{3}{5} \times \frac{2}{3}) \times \frac{- 6}{7}$
(d) $(\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5} = (\frac{- 6}{7} \times \frac{2}{3}) \times \frac{3}{5}$

Question 6

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. For the given orthogonal matrix Q.
Q =

⎡ ⎢ ⎣ \begin{matrix} 3 / 7 & 2 / 7 & 6 / 7 - 6 / 7 & 3 / 7 & 2 / 7 2 / 7 & 6 / 7 & - 3 / 7 \end{matrix} ⎤ ⎥ ⎦

The inverse is

6 \times (7 \times 3) = (6 \times 7) \times 3

is an example of

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√6 > 3√7

The correct option is A True Given surds are not 'Like surds'. So, to compare them the powers have to be made equal. Now LCM of 2 and 3 is 6. ⟹ √6 = 612×33 = 6√63 = 6√216 -------------(1) 3√7 = 713×22 = 6√72 = 6√49 -------------(2) From (1) and (2), 6√216> 6√49 ∴ √6 > 3√7

$\sqrt{6} > \sqrt[3]{7}$