For what values of 'K' does the equation have one real solution?

$$x^2 +2Kx+4=0$$​

The quadratic equation is:

​$$x^2 + 2Kx + 4 = 0$$​​

A quadratic equation $$ax^2 + bx + c = 0$$​ has one real solution (or a repeated real root) when its discriminant (D) is equal to zero.

the discriminant D is given by:

​$$D = b^2 - 4ac$$​​

Identify a, b, and c from the equation:

a = 1

b = 2K

c = 4

Substitute these values into the discriminant formula:

​$$D = (2K)^2 - 4 \cdot 1 \cdot 4$$​​

​$$D = 4K^2 - 16$$​​

​$$4K^2 - 16 = 0$$​​

​$$4K^2 = 16$$​​

​$$K^2 = 4$$​​

​$$K = ±2$$​​

Thus, The values of K for which the equation has one real solution are K = 2 and K = -2.