Math.sqrt(x) 
(Maybe you want to review carefully on what we learned about calling/invoking methods: click here )
Answer:
We have learned how to invoke methods on objects in this webpage: click here )
The above method call DOES NOT FIT the syntax that we learned in that webpage....
In order to "fit" the above method call in the syntax of a methods on objects, we will have to make the following associations:
Math.sqrt(x) ^ ^ ^      + Paramter  + Method name + Object 
But Math is a class and NOT an object
Example Program: _{ }
A class method has a special form: it does not operate on an object
What we would write IF we follow the syntax of an ordinary method call:  The correct way to express the operation to take the square root of the number x 

double x = 4; double root = x.sqrt(); // Wrong 
double x = 4; double root = Math.sqrt(x); // Good 
However, a number is NOT an object.
Therefore, the square root function does not work on objects.
I will repeat the syntax used to call an instance method for the sake of contrast:
objectName.methodName( parameters ) Example: Rectangle box = new Rectangle(5, 15, 20, 30); box.translate(2, 3); 
Syntax that you need to use to call a class method is:
ClassName.methodName( parameters ) Example: Math.sqrt(4); 
Math.sqrt(4); 
makes it appears as if the sqrt() method is applied to an "object" called Math  because Math preceeds sqrt() just like as box preceeds translate() in the method call box.translate()
The difference between box.translate(..) and Math.sqrt(..) is:
The guy who designed the Java programming language was heavily influenced by C++.

The object is passed to an instant method by means of the implicit parameter !!! 
Method names without the keyword static before its name tells you that the method is an instance method