To solve this problem, we need to determine the number of players, $n$, in a chess tournament where each player plays against every other player exactly once. The total number of games played is 36.

In a round-robin tournament where every player plays against every other player once, the total number of games is given by the combination formula:

$\binom{n}{2} = \frac{n(n-1)}{2}$

Here, $\binom{n}{2}$ represents the number of ways to choose 2 players from $n$.

Given that the total number of games is 36, we can set up the equation:

$\frac{n(n-1)}{2} = 36$

Step 1: Solve for $n$

Multiply through by 2 to eliminate the fraction:

$n(n-1) = 72$

This expands to:

$n^2 - n - 72 = 0$

Step 2: Factorize the quadratic equation

We need two numbers whose product is $-72$ and sum is $-1$. These numbers are $-9$ and $8$. Thus, we can factorize as:

$(n - 9)(n + 8) = 0$

Step 3: Solve for $n$

From the factors, $n - 9 = 0$ or $n + 8 = 0$, so:

$n = 9 \quad \text{or} \quad n = -8$

Since the number of players cannot be negative, we conclude:

$n = 9$

Final Answer:

There are 9 players in the tournament.

Do you want me to explain any steps further or explore related problems? Here are 5 related questions:

1. What if the number of games was 45? How many players would there be?
2. How does the formula $\binom{n}{2}$ work, and why is it used here?
3. Can this method be applied to tournaments where each pair plays multiple times?
4. How can this be extended to include ties or draws in games?
5. What if additional rounds are added in the tournament?

Tip: Always check your work by substituting the result back into the original formula to ensure accuracy!