Rather than trying to guess and check, we can actually construct a counterexample to the statement.
So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:

So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let
be a rational number. That would mean,
, would be a/b squared. Taking the square root of it yields:

So our stronger statement is proven, and we know that the original claim is decisively false.