SOLUTION: I need to check these four problems, can anyone help? 3^sqrt(-27) sqrt(-3x)^2 (with the square inside the square root sign) sqrt(-4) 3^sqrt(-1/8)

Algebra ->  Radicals -> SOLUTION: I need to check these four problems, can anyone help? 3^sqrt(-27) sqrt(-3x)^2 (with the square inside the square root sign) sqrt(-4) 3^sqrt(-1/8)       Log On


   

Question 114539: I need to check these four problems, can anyone help?
3^sqrt(-27)
sqrt(-3x)^2
(with the square inside the square root sign)
sqrt(-4)
3^sqrt(-1/8)
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I need to check these four problems, can anyone help?
3^sqrt(-27)= -3
--------------------
sqrt(-3x)^2 = -3x
--------------------
sqrt(-4) = 2i
-----------------
3^sqrt(-1/8) = -1/2
==========================
Cheers,
Stan H.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
1.
3%5Esqrt%28-27%29=+3%5E%28sqrt%2827%29sqrt%28-1%29%29=+3%5E%283i%29
2.
sqrt%28%28-3x%29%5E2%29+=+-3x
3.
sqrt%28-4%29+=+sqrt%284%29%2Asqrt%28-1%29+=+2i…………. imaginary number i%2Ai+=+-1
4.
3%5Esqrt%28-%281%2F8%29%29……………….. apply rule sqrt%28a%2Fb%29+=+sqrt%28a%29%2Fsqrt%28b%29
where b cannot be equal to 0


3%5E%28sqrt%28-1%29%2Fsqrt%288%29%29=3%5E%28i%2Fsqrt%288%29%29………… sqrt%28-1%29=i


Since sqrt%288%29=2.8284271247461903=2.83, we will have:


3%5E%28sqrt%28-1%29%2Fsqrt%288%29%29+=+3%5E%28i%2F2.83%29…………