Question:
There is a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational. Then, $x$ can be
(a) $\sqrt{5}$
(b) $\sqrt{2}$
(c) $\sqrt[3]{2}$
(d) $\sqrt[4]{2}$
Solution:
(a) Let $x=\sqrt{5}$.
$x^{2}=(\sqrt{5})^{2}=5$, which is a rational number.
(b) Let $x=\sqrt{2}$.
$x^{2}=(\sqrt{2})^{2}=2$, which is a rational number.
(c) Let $x=\sqrt[3]{2}$
$x^{2}=(\sqrt[3]{2})^{2}=(2)^{\frac{2}{3}}$, which is an irrational number.
$x^{4}=(\sqrt[3]{2})^{4}=(2)^{\frac{4}{3}}$, which is also an irrational number.
(d) Let $x=\sqrt[4]{2}$
$x^{2}=(\sqrt[4]{2})^{2}=(2)^{\frac{2}{4}}=(2)^{\frac{1}{2}}$, which is an irrational number.
$x^{4}=(\sqrt[4]{2})^{4}=(2)^{\frac{4}{4}}=2$, which is a rational number.
$\therefore x$ can be $\sqrt[4]{2}$
Hence, the correct option is (d).