The square root of 27b^11
Can someone explain to me how to solve this?
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~ :)