Question

There is a number x such that x² is irrational but x⁴ is rational. Then, x can be

(a) $\sqrt{5}$

(b) $\sqrt{2}$

(c) $\sqrt[3]{2}$

(d) $\sqrt[4]{2}$

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Let}x=\sqrt{5}. \\ {x}^2={\left(\sqrt{5}\right)}^2=5,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \\ \left(\mathrm{b}\right)\mathrm{Let}x=\sqrt{2}. \\ {x}^2={\left(\sqrt{2}\right)}^2=2,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \\ \left(\mathrm{c}\right)\mathrm{Let}x=\sqrt[3]{2}. \\ {x}^2={\left(\sqrt[3]{2}\right)}^2={\left(2\right)}^{\frac{2}{3}},\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ {x}^4={\left(\sqrt[3]{2}\right)}^4={\left(2\right)}^{\frac{4}{3}},\mathrm{which}\mathrm{is}\mathrm{also}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \left(\mathrm{d}\right)\mathrm{Let}x=\sqrt[4]{2}. \\ {x}^2={\left(\sqrt[4]{2}\right)}^2={\left(2\right)}^{\frac{2}{4}}={\left(2\right)}^{\frac{1}{2}},\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ {x}^4={\left(\sqrt[4]{2}\right)}^4={\left(2\right)}^{\frac{4}{4}}=2,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \end{gathered}$$

$\therefore$ x can be $\sqrt[4]{2}$.

Hence, the correct option is (d).