Which of the following surds is greatest in magnitude $\sqrt[6]{17}, \sqrt{2}, \sqrt[12]{25}, \sqrt[3]{4}$

\sqrt[6]{17}

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\sqrt[12]{25}

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\sqrt[3]{4}

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\sqrt{2}

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Solution

The correct option is A $\sqrt[6]{17}$
L.C.M of (2,6,3,12) is 12
Then $\sqrt[6]{17} = \sqrt[12]{17^{2}} = \sqrt[12]{289}$
$\sqrt{2} = \sqrt[12]{2^{6}} = \sqrt[12]{64}$
$\sqrt[12]{25} = \sqrt[12]{25} = \sqrt[12]{25}$
$\sqrt[3]{4} = \sqrt[12]{4^{4}} = \sqrt[12]{256}$
Hence $\sqrt[12]{289} > \sqrt[12]{256} > \sqrt[12]{64} > \sqrt[12]{25}$
so $\sqrt[6]{17}$ is greatest.

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