Show that each of the following is irrational :

(i) $\Big(2 + \sqrt{5}\Big)^2$

(ii) $\Big(3 - \sqrt{3}\Big)^2$

(iii) $\Big(\sqrt{5} + \sqrt{3}\Big)^2$

(iv) $\dfrac{6}{\sqrt{3}}$

(i) Given,

$\Rightarrow \Big(2 + \sqrt{5}\Big)^2 \\[1em] \Rightarrow (2)^2 + (\sqrt{5})^2 + 2 \times 2 \times \sqrt{5} \\[1em] \Rightarrow 4 + 5 + 4\sqrt{5} \\[1em] \Rightarrow 9 + 4\sqrt{5}.$

9 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

4 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{5}$ is an irrational number as it is a square root of a non-perfect square i.e. 5.

The product of a rational number and an irrational number is always irrational. i.e. $4\sqrt{5}$

The sum of a rational number and an irrational number is always irrational. i.e. $9 + 4\sqrt{5}$

Hence, $\Big(2 + \sqrt{5}\Big)^2$ is an irrational number.

(ii) Given,

$\Rightarrow \Big(3 - \sqrt{3}\Big)^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{3})^2 - 2 \times 3 \times \sqrt{3} \\[1em] \Rightarrow 9 + 3 - 6 \sqrt{3} \\[1em] \Rightarrow 12 - 6\sqrt{3} \\[1em]$

12 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

6 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The product of a rational number and an irrational number is always irrational. i.e. $6\sqrt{3}$

The difference between a rational number and an irrational number is always irrational. i.e. $12 - 6\sqrt{3}$.

Hence, $\Big(3 - \sqrt{3}\Big)^2$ is an irrational number.

(iii) Given,

$\Rightarrow \Big(\sqrt{5} + \sqrt{3}\Big)^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 + 2\sqrt{15} \\[1em] \Rightarrow 8 + 2\sqrt{15} \\[1em]$

8 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

2 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{15}$ is an irrational number as it is a square root of a non-perfect square i.e. 15.

The product of a rational number and an irrational number is always irrational. i.e. $2\sqrt{15}$

The sum of a rational number and an irrational number is always irrational. i.e. $8 + 2\sqrt{15}$

Hence, $\Big(\sqrt{5} + \sqrt{3}\Big)^2$ is an irrational number.

(iv) Given,

$\dfrac{6}{\sqrt{3}}$

6 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The division of a rational number and an irrational number is always irrational. i.e. $\dfrac{6}{\sqrt{3}}$

Hence, $\dfrac{6}{\sqrt{3}}$ is an irrational number.