Question

Compare the following pair of surds:

$\sqrt{3},\sqrt[4]{410}$

• A. $\sqrt{4}<\sqrt[4]{410}$
• B. $\sqrt{3}<\sqrt[3]{310}$
• C. $\sqrt{3}<\sqrt[4]{45}$
• D. $\sqrt{3}<\sqrt[4]{410}$

Answer 1

Answer: (D)

$\sqrt{3}<\sqrt[4]{410}$

$\sqrt{3},\sqrt[4]{410}$
To compare, the order of the surds must be same.
$=>3^{\frac{1}{2}},{10}^{\frac{1}{4}}$
$=>3^{\frac{2}{4}},{10}^{\frac{1}{4}}$
$=>\sqrt[4]{43^2},\sqrt[4]{410}$
$=>\sqrt[4]{49},\sqrt[4]{410}$
Now,$9<10$
$=>\sqrt[4]{49}<\sqrt[4]{410}$
$=>\sqrt{3}<\sqrt[4]{410}$