How is this possible/ what does it mean?

[Edi]

I am applying this to the General relativity .. stuff..
So, basically this [ http://www.wolframalpha.com/input/?i=(2/+sqrt(1-+400000000^2/300000000^2)) ] equation shows that it is pretty impossible to fly faster than the speed of light in vacuum, because the you have to deal with sqrt of negative numbers and that's just bad. Ok, fine. What else is new?

Oh, i know! This!: what if I take everything in the equation to the power of 4? Like this: [ http://www.wolframalpha.com/input/?i=(2^4/+sqrt(1-+400000000^2/300000000^2)^4) ]
of course, the result is in power of 4 too, but! - there is no negative numbers and problems!

This is a equation for mass increase while traveling at velocity near (faster? ) the speed of light in vacuum (c).
starting mass is 2(kg)
The resulting mass from traveling at 1.33 (?) c is ~26.45^1/4 .. well.. this : [ http://www.wolframalpha.com/input/?i=(2^4/+sqrt(1-+400000000^2/300000000^2)^4)^1/4 ]


Flying faster than c? No problem! ... except that there is a problem! :( - at least form my, not-that-much-educated, point of view.

 

[Mute]

You're just performing some mathematical manipulations that don't have any physical relevance. In particular, you're raising an imaginary number to the fourth power and taking the 4th root to get a real number. What you perhaps don't know is that when you take the nth root of a number, there are actually n solutions.

For example, take the number i. You know i^4 = 1. However, 1^(1/4) has four solutions: +1, -1, +i, -i.

So, at speeds faster than c you get an imaginary answer for $\gamma m_0$, but taking that to the fourth power and then taking the fourth root does not give you a physical answer that's real. The correct root is still the imaginary root you started with.