Question

what is the greatest possible integer value of x for which \\(\sqrt{x - 5}\\) is an imaginary number?
options: 3, 4, 5, 6

Explanation:

Step1: Recall when square root is imaginary

A square root \(\sqrt{a}\) is imaginary when \(a < 0\). So for \(\sqrt{x - 5}\) to be imaginary, we need \(x - 5 < 0\).

Step2: Solve the inequality

Solve \(x - 5 < 0\) by adding 5 to both sides: \(x < 5\).

Step3: Find the greatest integer x

We need the greatest integer \(x\) such that \(x < 5\). The integers less than 5 are 4, 3, 2, ..., so the greatest one is 4.

Answer:

4