If the quadratic equation, px² - $2\sqrt{5}$ px + 15 = 0 has two equal roots, then the value of p is:

1. 0

2. 3

3. 6

4. both 0 and 3

Comparing px² - $2\sqrt{5}$ px + 15 = 0 with ax² + bx + c = 0 we get,

a = p, b = $-2\sqrt{5}$ p and c = 15.

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ ($-2\sqrt{5}$ p)² - 4(p)(15) = 0

⇒ 20p² - 60p = 0

⇒ 20p(p - 3) = 0

⇒ 20p = 0 or (p - 3) = 0      [Using Zero-product rule]

⇒ p = 0 or p = 3.

Since p = 0 would make the equation no longer quadratic, we take:

p = 3

Hence, option 2 is the correct option.