What are the two consecutive negative integers whose product is 600, and what is the lesser integer?

To find two consecutive negative integers whose product equals 600, we can start by defining the two integers. Let the first integer be x. Since they are consecutive, the second integer will be x + 1. Given that the product of these integers is 600, we can set up the equation:

x * (x + 1) = 600

This simplifies to:

x² + x = 600

To rearrange this into a standard quadratic equation, we subtract 600 from both sides:

x² + x – 600 = 0

Next, we can use the quadratic formula to solve for x:

x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}

In our equation, a = 1, b = 1, and c = -600. Plugging these values into the formula gives:

x = \frac{-1 \pm \sqrt{1^2 – 4 \cdot 1 \cdot (-600)}}{2 \cdot 1}

Now, calculate the discriminant:

1 + 2400 = 2401

Now taking the square root:

\sqrt{2401} = 49

Now, substituting back we have:

x = \frac{-1 \pm 49}{2}

This will give us two potential solutions:

1. x = \frac{48}{2} = 24

2. x = \frac{-50}{2} = -25

Since we are looking for negative integers, we will take x = -25. The consecutive integers are:

-25 and -24.

Thus, the lesser integer is: -25.