Cosnider px² – 2√5 x + 15 = 0,

Here a = p , b = 2√5 p , c = 15

For real roots, D = 0

So,

b² – 4ac = 0

(2√5 p)² – 4p(15) = 0

20p² – 60p = 0

20p(p – 3) = 0

p = 3, p = 0

But p = 0 is not possible.

∴ p = 3.