If a is an integer, for how many integral values of n can the quadratic equation x2 – (2a + 3)x + 4n = 0, have real and equal roots for x?

The quadratic equation is given as x^2 - (2a + 3)x + 4n = 0.

For a quadratic equation ax^2 + bx + c = 0 to have real and equal roots, the discriminant (b^2 - 4ac) must be equal to zero.

In this case, a = 1, b = -(2a + 3), and c = 4n.

So, we have [-(2a + 3)]^2 - 414n = 0.

Solving this equation gives us:

(2a + 3)^2 = 16n.

Since a is an integer, (2a + 3)^2 is a perfect square. Let's denote it as m^2, where m is an integer.

So, we have m^2 = 16n, which implies n = m^2 / 16.

Since n is an integer, m^2 must be a multiple of 16. The possible values of m are ±4, ±8, ±12, ±16, ±20, ..., ±4k (where k is an integer).

So, there are infinite integral values of n for which the given quadratic equation can have real and equal roots for x.