How do you simplify #sqrt((2g^3)/(5z))#?

1 Answer
Ken C.
Sep 22, 2015

#(sqrt(10gz)*g)/(5z)#

Explanation:

First, begin by "distributing" the square root sign to the numerator and denominator:
#sqrt(2g^3)/(sqrt(5z)#
Now simplify the top:
#(sqrt(2)sqrt(g^3))/sqrt(5z)# (because #sqrt(ab) = sqrt(a)sqrt(b)#)

#(sqrt(2)*gsqrt(g))/sqrt(5z)# (for example, #sqrt(2^3) = sqrt(8) = 2sqrt(2)#)

Since it isn't proper to have a square root in the denominator, we take it out by multiplying by #sqrt(5z)/sqrt(5z)#.

#(sqrt(2)*gsqrt(g))/sqrt(5z)*sqrt(5z)/sqrt(5z)#

#(sqrt(2)*gsqrt(g)*sqrt(5z))/(5z)# (#sqrt(5z)*sqrt(5z) = 5z#)

We can now finally collect all of our radicals (square roots) into one neat sign:

#(sqrt(10zg)*g)/(5z)# (remember that #sqrt(a)sqrt(b)sqrt(c) = sqrt(abc)#)

And that is the final answer.

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