Final answer:
There are multiple real values of x for which the expression \(\sqrt{120} - \sqrt{x}\) is an integer, as any perfect square less than 120 can be subtracted from 120 to yield an integer. However, since none of the provided options correctly state 'multiple' or an actual count, the closest choice would be 'None'. the closest answer choice is None, as the other options suggest a specific number of values.
Step-by-step explanation:
To determine how many real values of x make the expression \(\sqrt{120} - \sqrt{x}\) an integer, we need to consider the properties of square roots and integers. Since \(\sqrt{120}\) is a non-integer square root (between the square roots of \(11^2 = 121\) and \(10^2 = 100\)), subtracting another square root can only result in an integer if the second square root is also non-integer, and their difference is an integer.
Likewise, \(\sqrt{120}\) can be approximated to \(10.95\). The square roots of perfect squares closest to \(120\) are \(\sqrt{121} = 11\) and \(\sqrt{100} = 10\). While \(11 - \sqrt{x}\) can be an integer, for \(x = 0, 1, 4, 9, 16, ...\), all the way up to \(x = 120\) (the square root of \(x\) must be less than \(11\) to not result in a negative integer), the difference between \(\sqrt{120}\) and \(\sqrt{x}\) can be one of these integers.
Therefore, there will be multiple values of x where the expression yields an integer. Since this covers a range of values and not a fixed count, the closest answer choice is None, as the other options suggest a specific number of values.