what is the greatest possible integer value of x for which $sqrt{x - 5}$ is an imaginary number?\n3\n4\n5\n6

Explanation:

Step1: Recall the condition for imaginary number

For $\sqrt{a}$ to be an imaginary number, $a<0$. So for $\sqrt{x - 5}$ to be imaginary, $x-5<0$.

Step2: Solve the inequality

$x-5<0$ implies $x < 5$.

Step3: Find the greatest integer value

The greatest integer less than 5 is 4.

Answer:

B. 4