Question

Which of the following are irrational numbers?
(i) $\sqrt{2+\sqrt{3}}$
(ii) $\sqrt{4+\sqrt{25}}$
(iii) $\sqrt[3]{35+\sqrt{7}}$
(iv) $\sqrt{8-\sqrt[3]{38}}$.

• A. (ii), (iii) and (iv)
• B. (i), (ii) and (iv)
• C. (i), (ii) and (iii)
• D. (i), (iii) and (iv)

Answer 1

The correct option is D (i), (iii) and (iv)

Option $\left(i\right)$

$\sqrt{3}$ is irrational, so $\sqrt{2+\sqrt{3}}$ is also irrational.

Option $\left(ii\right)$

$\sqrt{25}=5$, so we get $\sqrt{4+5}=\sqrt{9}=3$, which is rational.

Option $\left(iii\right)$

$\sqrt{7}$ is irrational, so $\sqrt[3]{35+\sqrt{7}}$ is also irrational.

Option $\left(iv\right)$

$\sqrt[3]{38}=2$, so we get $\sqrt{8-2}=\sqrt{6}$, which is irrational.

$\therefore$ $\left(i\right)$, $\left(iii\right)$ and $\left(iv\right)$ are irrational.

That is, option $D$ is correct.