lowergm

We can solve this problem using the discriminant of the quadratic equation. For a quadratic equation of the form ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. If the discriminant is positive, the quadratic equation has two distinct real roots. If the discriminant is zero, the quadratic equation has one real root. If the discriminant is negative, the quadratic equation has no real roots.

In this problem, we are given the quadratic equation x^2 - (a + 1)x + a = 0, and we want to find the product of all values of a for which the equation has exactly one real solution.

Using the quadratic formula, we can solve for x:

x = [(a + 1) ± sqrt((a + 1)^2 - 4a)] / 2

The equation has exactly one real solution when the discriminant is zero:

(a + 1)^2 - 4a = 0

Simplifying this equation, we get:

a^2 - 2a + 1 = 0