In each of the following, determine whether the given numbers are solutions of the given equations or not:

(i) $x^2 - 3\sqrt{3}x + 6 = 0; \sqrt{3}, \space -2\sqrt{3}$

(ii) $x^2 - \sqrt{2}x - 4 = 0; -\sqrt{2}, \space 2\sqrt{2}$

(i) Given,

$x^2 - 3\sqrt{3}x + 6 = 0$

Substituting x = $\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (\sqrt{3})^2 - 3\sqrt{3} \times \sqrt{3} + 6 \\[0.5em] = 3 - 9 + 6 \\[0.5em] = 9 - 9 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $\sqrt{3}$ is a solution of the given equation

Substituting x = $-2\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (-2\sqrt{3})^2 - 3\sqrt{3} \times -2\sqrt{3} + 6 \\[0.5em] = 12 + 18 + 6 \\[0.5em] = 36$

≠ $\text{RHS}$

∴ $-2\sqrt{3}$ is not a solution of the given equation

Hence, $\sqrt{3}$ is a solution of the given equation but $-2\sqrt{3}$ is not.

(ii) Given,

$x^2 - \sqrt{2}x - 4 = 0$

Substituting x = $-\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (-\sqrt{2})^2 - \sqrt{2} \times (-\sqrt{2}) - 4 \\[0.5em] = 2 + 2 - 4 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $-\sqrt{2}$ is a solution of the given equation

Substituting x = $2\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (2\sqrt{2})^2 - \sqrt{2} \times 2\sqrt{2} - 4 \\[0.5em] = 8 - 4 - 4 \\[0.5em] = 8 - 8 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $2\sqrt{2}$ is a solution of the given equation

Hence, $-\sqrt{2}$ and $2\sqrt{2}$ both are solutions of the given equation.