The product of two consecutive positive integers is 812. What is the value of the lesser integer?

To find the lesser integer in the pair of consecutive integers whose product is 812, we can denote the two integers as x and x + 1. The equation representing their product is:

x * (x + 1) = 812

Expanding this, we get:

x² + x – 812 = 0

This is a quadratic equation in the standard form ax² + bx + c = 0. Here, a = 1, b = 1, and c = -812.

We can solve for x using the quadratic formula:

x = (-b ± √(b² – 4ac)) / 2a

Substituting the values:

x = (-1 ± √(1² – 4 * 1 * -812)) / (2 * 1)

Calculating the discriminant:

1 + 3248 = 3249

So, we proceed with:

x = (-1 ± √3249) / 2

We find that √3249 = 57, thus:

x = (-1 + 57) / 2

This gives us:

x = 28

Now, we can conclude that the two consecutive integers are 28 and 29. Therefore, the value of the lesser integer is:

28