Asked by Please Help
The square root of 27b^11
Can someone explain to me how to solve this?
Can someone explain to me how to solve this?
Answers
Answered by
Jai
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~ :)
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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