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Comparison of irrational numbers - class-IX
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Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$
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A
$\sqrt[8]{80} < \sqrt[4]{40}$
💡 Explanation:
$\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
$80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$