Note: For a quadratic equation, ax² + bx + c = 0, we have D = b² – 4ac.

If D = 0, then the roots of the quadratic equation are equal.

Therefore, (p+1)x² - 6(p+1)x + 3(p + 9) = 0 will have equal roots when, 

⇒ D = 0  

⇒ b² – 4ac = 0

⇒ b² = 4ac 

Here, b = -6(p+1),

         a = (p+1)

and,   c = 3(p+9)
⇒{-6(p + 1)}² = 4×(p + 1)×3(p + 9) 

⇒ 36(p+1)(p+1) = 12(p + 1)(p + 9) 
⇒ 3(p+1)=(p + 9)

 ⇒ 3p + 3 - p - 9 = 0

⇒ 2p - 6 = 0

⇒ p = 6/2

⇒ p = 3 

Thus, the value of p is 3

Now, putting the value of p in (p+1)x² - 6(p+1)x + 3(p + 9) = 0, we get, 

⇒ 4x² - 24x + 36 = 0

On taking 4 common, we get, 

⇒ x² – 6x + 9 = 0

⇒ (x - 3)² = 0

⇒ x = 3

Thus, the root of the given equation is  x = 3