SOLUTION: The question is how do I find (in interval notation) the values of x for which this expressio is real. Squareroote of (15-5x). The anwswer seems to be any number beyond 3 wil

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Question 607126: The question is how do I find (in interval notation) the values of x for which this expressio is real.
Squareroote of (15-5x).
The anwswer seems to be any number beyond 3 will produce a real number. so the answer should be (3, infinity) correct?
Found 2 solutions by LisaJ, ewatrrr:
Answer by LisaJ(11) About Me  (Show Source):
You can put this solution on YOUR website!
That is almost correct, because x=3 is the cutoff point. But this expression is real when x is less than or equal to 3, not when x is greater than 3. You can see that by trying a larger number, such as x = 10. Then (15-5x) is negative, and the square root is not real.
On the other hand, when x = 0, (15-5x) = 15, and the square root is real.
So the answer is (-infinity, 3).

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

values of x for which this expression is real.

sqrt%2815-5x%29

Expression in real when (15-5x) ≥ 0 or (−∞,3]

Note:

 15-5x ≥ 0

  15 ≥ 5x

   3 ≥ x