Posted by **Please Help** on Wednesday, May 4, 2011 at 9:26pm.

The square root of 27b^11

Can someone explain to me how to solve this?

- Math -
**Jai**, Wednesday, May 4, 2011 at 9:41pm
first, we factor the radicand (the term inside the radical sign or the squreroot):

sqrt(27*b^11)

sqrt[ (3*3*3) (b*b^10) ]

then we separate the perfect squares:

sqrt[ 3*(3^2) b*((b^5)^2) ]

and we take the squarerrot of the perfect squres. the squareroot of 3^2 is 3, while (b^5)^2 is b^5, then we can take them out of the sqrt, leaving only the 3 and b inside:

3b^5 * sqrt(3b)

hope this helps~ :)

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