Question

Arrange the following in descending order:
$\sqrt[3]{34},\sqrt{2},\sqrt[6]{63},\sqrt[4]{45}$.

• A. $\sqrt[3]{34}>\sqrt[4]{45}>\sqrt{2}>\sqrt[6]{63}$
• B. $\sqrt[4]{45}>\sqrt[3]{34}>\sqrt[6]{63}>\sqrt{2}$
• C. $\sqrt{2}>\sqrt[6]{63}>\sqrt[3]{34}>\sqrt[4]{45}$
• D. $\sqrt[6]{63}>\sqrt[4]{45}>\sqrt[3]{34}>\sqrt{2}$

Answer 1

The correct option is A $\sqrt[3]{34}>\sqrt[4]{45}>\sqrt{2}>\sqrt[6]{63}$

Given that,
$\sqrt[3]{34},\sqrt{2},\sqrt[6]{63},\sqrt[4]{45}$
$(4{)}^{\frac{1}{3}},(2{)}^{\frac{1}{2}},(3{)}^{\frac{1}{6}},(5{)}^{\frac{1}{4}}$
LCM of $3,2,6$ and $4$ is $12$.
$\sqrt[3]{34}=\sqrt[3\times 4]{3\times 44^4}=\sqrt[12]{12256}$
$\sqrt{2}=\sqrt[2\times 6]{2\times 62^6}=\sqrt[12]{1264}$
$\sqrt[4]{45}=\sqrt[4\times 2]{4\times 25^3}=\sqrt[12]{12125}$
$\sqrt[6]{63}=\sqrt[6\times 2]{6\times 23^2}=\sqrt[12]{129}$
Hence, In descending order they are
$\sqrt[3]{34}>\sqrt[4]{45}>\sqrt{2}>\sqrt[6]{63}$.