Question

Arrange the following in ascending order of magnitude:

$\sqrt[4]{43},\sqrt[6]{610},\sqrt[12]{1225}$

• A. $\sqrt[12]{1225}>\sqrt[4]{43}>\sqrt[6]{610}$
• B. $\sqrt[12]{1225}<\sqrt[4]{43}>\sqrt[6]{610}$
• C. $\sqrt[12]{1225}>\sqrt[4]{43}<\sqrt[6]{610}$
• D. $\sqrt[12]{1225}<\sqrt[4]{43}<\sqrt[6]{610}$

Answer 1

The correct option is D $\sqrt[12]{1225}<\sqrt[4]{43}<\sqrt[6]{610}$

$\sqrt[4]{43},\sqrt[6]{610},\sqrt[12]{1225}$

To compare, the order of the surds must be same.

$=>3^{\frac{1}{4}},{10}^{\frac{1}{6}},{25}^{\frac{1}{12}}$

$=>3^{\frac{3}{12}},{10}^{\frac{2}{12}},{25}^{\frac{1}{12}}$

$=>\sqrt[12]{123^3},\sqrt[12]{12{10}^2},\sqrt[12]{1225}$

$=>\sqrt[12]{1227},\sqrt[12]{12100},\sqrt[12]{1225}$

Therefore,
$=>25<27<100$

$=>,\sqrt[12]{1225}<\sqrt[12]{1227}<\sqrt[12]{12100}$

$=>\sqrt[12]{1225}<\sqrt[4]{43}<\sqrt[6]{610}$