Examine whether the following numbers are rational or irrational :

(i) $(3- \sqrt5)^2$

(ii) $(7 - \sqrt7)(7 + \sqrt7)$

(iii) $(2\sqrt3 + 3\sqrt2)^2$

(iv) $(2\sqrt3 - 3\sqrt2)(2\sqrt3 + 3\sqrt2)$

(v) $5\sqrt3 \times \sqrt{12}$

(i)

$(3 - \sqrt5)^2 = 3^2 - 2 \times 3 \times \sqrt5 + (\sqrt5)^2\\[1em] = 9 - 6\sqrt5 + 5\\[1em] = 14 - 6\sqrt5$

Since, 14 is rational, $6\sqrt5$ is irrational and we know that the sum of a rational and an irrational number is always irrational.

∴ (14 - $6\sqrt5$) is an irrational number.

Hence, $(3- \sqrt5)^2$ is an irrational number.

(ii)

$(7 - \sqrt7)(7 + \sqrt7) = 7^2 - (\sqrt7)^2\\[1em] = 49 - 7\\[1em] = 42$

∴ 42 is a rational number.

Hence, $(7 - \sqrt7)(7 + \sqrt7)$ is a rational number.

(iii)

$(2\sqrt3 + 3\sqrt2)^2 = (2\sqrt3)^2 + 2 \times 2\sqrt3 \times 3\sqrt2 + (3\sqrt2)^2\\[1em] = 12 + 12\sqrt6 + 18\\[1em] = 30 + 12\sqrt6\\[1em]$

Since, 30 is rational, $12\sqrt6$ is irrational and we know that the sum of a rational and an irrational number is always irrational.

∴ (30 + $12\sqrt6$) is an irrational number.

Hence, $(2\sqrt3 + 3\sqrt2)^2$ is an irrational number.

(iv)

$(2\sqrt3 - 3\sqrt2)(2\sqrt3 + 3\sqrt2) = (2\sqrt3)^2 - (3\sqrt2)^2\\[1em] = 4 \times 3 - 9 \times 2\\[1em] = 12 - 18\\[1em] = -6$

∴ -6 is a rational number.

Hence, $(2\sqrt3 - 3\sqrt2)(2\sqrt3 + 3\sqrt2)$ is a rational number.

(v)

$5\sqrt3 \times \sqrt{12} = 5 \sqrt{3 \times 12}\\[1em] = 5 \sqrt{36}\\[1em] = 5 \times 6\\[1em] = 30$

∴ 30 is a rational number.

Hence, $5\sqrt3 \times \sqrt{12}$ is a rational number.