Show that one root of the quadratic equation x² + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Substituting x = -3 in x² + (3 - 2a)x - 6a = 0,

⇒ (-3)² + (3 - 2a)(-3) - 6a = 0

⇒ 9 - 9 + 6a - 6a = 0

⇒ 0 = 0.

Hence, -3 is one root of the quadratic equation.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$= \dfrac{-(3 - 2a) \pm \sqrt{(3 - 2a)^2 - 4(1)(-6a)}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{9 + 4a^2 - 12a + 24a}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{(2a - 3) + (2a + 3)}{2} \text{ or } \dfrac{(2a - 3) - (2a + 3)}{2} \\[1em] = \dfrac{4a}{2} \text{ or } \dfrac{-6}{2} \\[1em] = 2a \text{ or } -3.$

Hence, the other root is 2a.