Find the values of m for which equation 3x² + mx + 2 = 0 has equal roots. Also, find the roots of the given equation.

When equation has equal roots, D = 0.

∴ b² - 4ac = 0

Comparing 3x² + mx + 2 = 0 with ax² + bx + c = 0 we get,

a = 3, b = m and c = 2.

⇒ m² - 4(3)(2) = 0

⇒ m² - 24 = 0

⇒ m² = 24

⇒ m = $\sqrt{24} = \pm 2 \sqrt{6}$

Considering, m = $2\sqrt{6}$

⇒ 3x² + mx + 2 = 0

3x² + $2\sqrt{6}$x + 2 = 0

3x² + $\sqrt{6}x + \sqrt{6}x + 2$ = 0

$(\sqrt{3}x + \sqrt{2})^2$ = 0

$(\sqrt{3}x + \sqrt{2})$ = 0

x = $-\dfrac{\sqrt{2}}{\sqrt{3}}$.

Rationalising,

$x = -\dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = - \dfrac{\sqrt{6}}{3}$.

Considering, m = -$2\sqrt{6}$

⇒ 3x² + mx + 2 = 0

⇒ 3x² - $2\sqrt{6}$x + 2 = 0

⇒ $(\sqrt{3}x - \sqrt{2})^2$ = 0

⇒ $(\sqrt{3}x - \sqrt{2})$ = 0

⇒ x = $\dfrac{\sqrt{2}}{\sqrt{3}}$.

Rationalising,

$x = \dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{6}}{3}$.

Hence, m = $\pm 2\sqrt{6}$ and roots = $\pm \dfrac{\sqrt{6}}{3}$.