(iii) $\sqrt{2}, \sqrt[3]{5}$

Open in App

Solution

The orders of $\sqrt{2} and \sqrt[3]{5}$ are 2 and 3.
The LCM of 2 and 3 is 6.
So, we will make the orders of both the surds 6.
$\sqrt{2} = \sqrt[6]{2^{3}} = \sqrt[6]{8} and \sqrt[3]{5} = \sqrt[6]{5^{2}} = \sqrt[6]{25}$
Hence, the order is same.
Since 8 < 25, $\sqrt{2} < \sqrt[3]{5}$ .

Suggest Corrections

Similar questions

Question 124 (iii)

Evaluate the following, using distributive property.
$53 \times (- 9) - (- 109) \times 53$

Using the identities of algebraic expressions, Find :

$(i) 102^{2}$

$(i i) 53^{2}$

$(i i i) 1001^{2}$ [3 MARKS]

2^{10}

5^{3}

(- 7)^{3}

8^{1}

\sqrt{50}

\sqrt[3]{5}

\sqrt{31^{\frac{1}{3}}}

^{3} \sqrt{(9)^{2}}

\sqrt[4]{\frac{31}{50}}

Join BYJU'S Learning Program