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

(i) $x^2 - 5x + 6 = 0; 2, -3$

(ii) $3x^2 - 13x - 10 = 0 ; 5, -\dfrac{2}{3}$

(i) Given,

$x^2 - 5x + 6 = 0$

Substituting x = 2 in the LHS of the given equation, we get:

$\text{LHS} = 2^2 - 5 \times 2 + 6 \\[0.5em] = 4 - 10 + 6 \\[0.5em] = 10 - 10 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 2 is a root of the given equation.

Substituting x = -3 in the LHS of the given equation, we get:

$\text{LHS} = (-3)^2 - 5 \times (-3) + 6 \\[0.5em] = 9 + 15 + 6 \\[0.5em] = 30$

≠ $\text{RHS}$

∴ 3 is not a root of the given equation.

Hence, 2 is a root but -3 is not a root of the given equation.

(ii) Given,

$3x^2 - 13x - 10 = 0$

Substituting x = 5 in the LHS of the given equation, we get:

$\text{LHS} = 3 \times (5)^2 - 13 \times 5 - 10 \\[0.5em] = 3 \times 25 - 65 -10 \\[0.5em] = 75 - 75 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 5 is a root of the equation

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

$\text{LHS} = 3 \times \Big(-\dfrac{2}{3}\Big)^2 - 13 \times \Big( -\dfrac{2}{3} \Big) - 10 \\[1em] = 3 \times \dfrac{4}{9} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{4}{3} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{30}{3} - 10 \\[1em] = 10 - 10 \\[1em] = 0 \\[1em] = \text{RHS}$

∴ $-\dfrac{2}{3}$ is a root of the equation

Hence, both 5 and $-\dfrac{2}{3}$ are roots of the given equation.