Question

Arrange the following surds in ascending order of their magnitudes: $\sqrt{5},\sqrt[3]{311},2\sqrt[6]{63}$

• A. $\sqrt[3]{311}>\sqrt{5}<2\sqrt[6]{63}$
• B. $\sqrt[3]{311}<\sqrt{5}<2\sqrt[6]{63}$
• C. $\sqrt[3]{311}>\sqrt{5}>2\sqrt[6]{63}$
• D. $\sqrt[3]{311}<\sqrt{5}>2\sqrt[6]{63}$

Answer 1

The correct option is B $\sqrt[3]{311}<\sqrt{5}<2\sqrt[6]{63}$

$\sqrt{5}=5^{1/2}$

$\sqrt[3]{311}={11}^{1/3}$

$2\sqrt[6]{63}=\sqrt[6]{612}={12}^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5}=5^{\frac{1}{2}\times \frac{6}{6}}=\sqrt[12]{125^6}=\sqrt[12]{1215625}$

$\sqrt[3]{311}={11}^{\frac{1}{3}\times \frac{4}{4}}=\sqrt[12]{12{11}^4}=\sqrt[12]{1214641}$

$2\sqrt[6]{63}=\sqrt[6]{612}={12}^{\frac{1}{6}\times \frac{2}{2}}=\sqrt[12]{12{12}^2}=\sqrt[12]{12144}$

Hence, the Ascending order is $2\sqrt[6]{63},\sqrt[3]{311},\sqrt{5}$