Question

The correct descending order of the following surds is
$\sqrt[3]{32}$, $\sqrt{3}$, $\sqrt{4}$, $\sqrt{5}$

• A. $\sqrt{3}$ > $\sqrt{4}$ > $\sqrt{5}$ > $\sqrt[3]{32}$
• B. $\sqrt{4}$ > $\sqrt{5}$ > $\sqrt[3]{32}$ > $\sqrt{3}$
• C. $\sqrt{5}$ > $\sqrt{4}$ > $\sqrt{3}$ > $\sqrt[3]{32}$
• D. $\sqrt[3]{32}$ > $\sqrt{4}$ > $\sqrt{3}$ > $\sqrt{5}$

Answer 1

Answer: (C)

$\sqrt{5}$ > $\sqrt{4}$ > $\sqrt{3}$ > $\sqrt[3]{32}$

LCM of $3$ and $2$ is $6$.
Therefore, multiplying the index of all numbers by $6$.
${\sqrt[3]{32}}^6=2^{\frac{6}{3}}=2^2=4$

${\sqrt{3}}^6=3^{\frac{6}{2}}=3^3=27$

${\sqrt{4}}^6=4^{\frac{6}{2}}=4^3=64$

${\sqrt{5}}^6=5^{\frac{6}{2}}=5^3=125$

$\therefore 125>64>27>4$

So, option $C$ is correct.