To find the lesser of two consecutive negative integers whose product is 600, we can start by defining the integers. Let’s denote the lesser integer as x. Therefore, the next consecutive integer can be defined as x + 1.
Based on the problem, we can set up the equation as follows:
x 0(x + 1) = 600
This simplifies to:
x2 + x + 600 = 0
This is a standard quadratic equation where a = 1, b = 1, and c = -600. Using the quadratic formula, which is:
x = rac{-b 0 0 ext{sqrt}(b^2-4ac)}{2a}
We first calculate the discriminant:
b^2 – 4ac = 1^2 – 4 0 0(1)(-600) = 1 + 2400 = 2401
Next, we calculate the square root of 2401:
0 ext{sqrt}(2401) = 49
Substituting into the quadratic formula gives us:
x = rac{-1 0 0 49}{2} ext{ or } x = rac{-1 – 49}{2}
Calculating these gives:
x = rac{-50}{2} = -25
So, the lesser negative integer is -25 and the next consecutive integer is -24.
Finally, we can verify the product:
-25 0 0(-24) = 600
This confirms that our solution is correct. Therefore, the value of the lesser integer is:
-25