The product of two consecutive negative integers is 2550. What is the sum of the two integers?
We start out with the equation $n(n + 1) = 2550$. Expanding, we find $n^2 + n - 2550 = 0$. This factors to $(n - 50)(n + 51) = 0$, so $n = 50\text{ or }-51.$ Since $n$ must be negative, we have $n = -51$, hence our two integers are $n = -51$ and $n + 1= -50$, which add up to $\boxed{-101}$.
-101
Level 2
Algebra
Problem: The product of two consecutive negative integers is 2550. What is the sum of the two integers?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find two consecutive negative integers whose product is 2550, let the integers be represented as x and x + 1, where x is a negative integer. The equation can be set up as:

x * (x + 1) = 2550.

Expanding this gives:

x^2 + x = 2550.

Rearranging the equation leads to:

x^2 + x - 2550 = 0.

This is a quadratic equation in standard form. To solve for x, we can apply the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a,

where a = 1, b = 1, and c = -2550. Plugging in these values:

b^2 - 4ac = 1^2 - 4 * 1 * (-2550) = 1 + 10200 = 10201.

Now, we calculate the square root:

sqrt(10201) = 101.

Substituting back into the quadratic formula gives:

x = [-1 ± 101] / 2.

Calculating the two possible values for x:

1. x = (-1 + 101) / 2 = 100 / 2 = 50 (not negative).
2. x = (-1 - 101) / 2 = -102 / 2 = -51 (negative).

Thus, the two consecutive negative integers are -51 and -50. To find their sum:

-51 + (-50) = -101.

Therefore, the final answer is -101.<|END_OF_LATENT|> The final answer is $-101$.