given any real number,k, the square root of the square root of (-k)*4 is always equal to x?

Question

do i need the knowledge of squsaring a neg number with imaginary numbers???
prove or disprove

x is real number

Answer 1

Your question makes no sense. What is x?
Also the asterick (*) usually means multiply. It looks that you intend it to mean exponentiation for which the character (^) is generally used.

So I think your question shoud read "given any real number,k, the square root of the square root of (-k)^4 is always equal to k?".

If k =<0, the statement is always true with the exception that ther are 3 other roots, two of which are imaginary.
If k is positive then we have , for example, (-2)^4 = 16
16^1/4 = 2 or -2. So yes it can = -2 but it can also = -2. There are also two additional complex roots. ( a 4th degree equation always has 4 roots, 2 or 4 of which could be complex imaginary numbers.)
Your statement could be written [(-k)^4]^1/4 = (-k) ^1 = -k, which might convince some people your conjecture is true.

But it is only true with some conditions as outlined above.

Answer 2

2i. underroot(k)...... i = underroot (-1) = imaginary