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Understanding what a real number is forms the bedrock of solving math problems that involve square roots and inequalities. Real numbers include all the numbers on the number line. This encompasses rational numbers like fractions and integers, as well as irrational numbers, which cannot be expressed as a simple fraction. For example, \( \frac{1}{2} \) is rational, while \( \sqrt{2} \) is irrational but both are real numbers.

To determine if an expression involving a square root represents a real number, the value under the square root, known as the radicand, must be non-negative. This is because in the real number system, the square root of a negative number is not defined. The exercise asks for which values of \(x\) the square root \( \( \sqrt{2 x-6} \) \) is real. This directly translates to finding the range of \(x\) values for which the radicand \(2x-6\) is greater than or equal to zero.

A square root is a number that, when multiplied by itself, yields the original number. For instance, the square root of 25, denoted as \( \( \sqrt{25} \) \) is 5, because \( 5 \times 5 = 25 \). Square roots are ubiquitous in mathematics and are fundamental in solving various types of problems.

It's crucial to remember that the principal square root, which is the non-negative root, is generally considered when solving basic equations. In the provided exercise, we want to ensure that the expression under the square root \( \( \sqrt{2 x-6} \) \) is non-negative, as square roots of negative numbers are not real. The exercise effectively filters out any values that would result in an imaginary number, which is outside the scope of this problem.

Inequalities are mathematical expressions that show the relationship between two values where they are not equal. In the context of square roots, solving an inequality ensures the radicand is non-negative, ensuring the result is a real number.

To find the inequality solution for the exercise, we set up the inequality \( 2x - 6 \text{\geq} 0 \) to find the values of \(x\) that make \( \( \sqrt{2 x-6} \) \) real. By adding 6 to both sides and then dividing by 2, as shown in the step-by-step solution, we find \( x \text{\geq} 3 \). This final inequality shows the range of values for \(x\) that are solutions to the problem. Essentially, any real number 3 or greater will keep the radicand non-negative and thus maintain the reality of the square root function.