### Instruction: A triangle has three sides of the following side lengths: $7$, $10$, and $x^2$. What are all of the positive integer values of $x$ such that the triangle exists? Separate your answers using commas and express them in increasing order.### Response: I know that a triangle exists if and only if the sum of any two sides is greater than the third side.
This is called the triangle inequality.
So, I need to check three cases: $7 + 10 > x^2$, $7 + x^2 > 10$, and $10 + x^2 > 7$.
The first case simplifies to $x^2 < 17$, which means $x < \sqrt{17}$.
The second case simplifies to $x^2 > 3$, which means $x > \sqrt{3}$.
The third case simplifies to $x^2 > -3$, which is always true for any real number $x$.
Therefore, the triangle exists if and only if $\sqrt{3} < x < \sqrt{17}$.
But the problem asks for positive integer values of $x$, so I need to find the integers in this interval.
I know that $\sqrt{3}$ is about $1.7$ and $\sqrt{17}$ is about $4.1$.
So, the only integers in this interval are $2$, $3$, and $4$.
These are the possible values of $x$ that make the triangle exist.