Sqrt(-36a^3b^4)

[Casey]

Sqrt(-36a^3b^4) = 6|a|b^2i sqrt(3) ??
so I got that it would become
i sqrt(36a^3b^4)
but what happens next? How do you get a cubed/^4 out from a root?
I know that I already have the answer, but I need to show the work. Please help, I'm desperate!

 

[mmm4444bot]

Sqrt(-36a^3b^4) = 6|a|b^2i sqrt(3) …

Did you intend to type sqrt(a) instead of sqrt(3) ?

… How do you get a cubed/^4 out from a root …

I'm not familiar with the notation "cubed/^4". Can you explain what you mean by this ?

 

[Casey]

Re:

Sqrt(-36a^3b^4) = 6|a|b^2i sqrt(a)
I did, thanks, its sqrt(a)
… How do you get a ^3 and ^4 out from a (sqrt)?

 

[Deleted member 4993]

Re: Re:

Sqrt(-36a^3b^4) = 6|a|b^2i sqrt(a)
I did, thanks, its sqrt(a)
… How do you get a ^3 and ^4 out from a (sqrt)?

You factorize

sqrt(a[sup:1bon02gw]7[/sup:1bon02gw]) = sqrt(a[sup:1bon02gw]6[/sup:1bon02gw] * a) = sqrt(a[sup:1bon02gw]6[/sup:1bon02gw]) * sqrt(a) = a[sup:1bon02gw]3[/sup:1bon02gw] * sqrt(a)

 

[Casey]

Re: Re:

Sqrt(-36a^3b^4) = 6|a|b^2i sqrt(a)
I did, thanks, its sqrt(a)
… How do you get a ^3 and ^4 out from a (sqrt)?

You factorize

sqrt(a[sup:zdmvv4jb]7[/sup:zdmvv4jb]) = sqrt(a[sup:zdmvv4jb]6[/sup:zdmvv4jb] * a) = sqrt(a[sup:zdmvv4jb]6[/sup:zdmvv4jb]) * sqrt(a) = a[sup:zdmvv4jb]3[/sup:zdmvv4jb] * sqrt(a)

Okay so...
I think i get it...
But whats with the absolute value thing? I dont understand that part either.
Thanks so much for the help

 

[Denis]

Re: Re:

But whats with the absolute value thing? I dont understand that part either.

Sqrt(-36 a^3 b^4)
Hint: say a = -3 ; then a^3 = -3 * -3 = 9 * -3 = -27

What would the solution be if problem was stated like this:
Sqrt(-36 a^3 b^4) where a < 0 ?