Find the value of k for which x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0.

Thus, find the other root of the equation.

Substituting, x = 3 in (k + 2)x² - kx + 6 = 0 we get,

⇒ (k + 2)(3)² - 3k + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\dfrac{-24}{6}$

⇒ k = -4.

Substitute the value of k = -4 in (k + 2)x² - kx + 6 = 0 we get,

⇒ (-4 + 2)(x)² - (-4)x + 6 = 0

⇒ (-2)(x)² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2x² - 2x + 6x + 6 = 0

⇒ -2x(x + 1) + 6(x + 1) = 0

⇒ (x + 1)(-2x + 6) = 0

⇒ (x + 1) = 0 or (-2x + 6) = 0      [Using Zero-product rule]

⇒ x = -1 or -2x = -6

⇒ x = -1 or x = $\dfrac{-6}{-2}$

⇒ x = -1 or x = 3.

Hence, the value of k = -4 and the other root is -1.