What are the two consecutive positive integers whose product is 812, and what is the value of the lesser integer?

To find the two consecutive positive integers whose product is 812, we can define the integers as n (the lesser integer) and n + 1 (the greater integer). The equation representing their product can be expressed as:

n(n + 1) = 812

This expands to:

n² + n – 812 = 0

This is a quadratic equation, and we can solve for n using the quadratic formula:

n = (-b ± √(b² – 4ac)) / 2a

In our equation, a = 1, b = 1, and c = -812. Plugging in these values:

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

n = (-1 ± √(1 + 3248)) / 2

n = (-1 ± √3249) / 2

Calculating the square root:

√3249 = 57

So, we have:

n = (-1 ± 57) / 2

Considering the positive root:

n = (56) / 2 = 28

Thus, the lesser integer is 28

The two consecutive integers are therefore 28 and 29. To verify:

28 × 29 = 812

This confirms our solution. Hence, the value of the lesser integer is 28.