How do you simplify #sqrt(20y^5)#?

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Answer 1 · 2015-09-18T16:31:14

#sqrt(20y^5) = 2y^2sqrt(5y)#

We know we can only take perfect squares out of the root, so we have to find the square factors in 20 and in #y^5#

The latter is easy, we know that #y^5 = y^2*y^2*y#, so we can take out two #y#s

#sqrt(20y^5) = y^2sqrt(20y)#

20 is more complicated, but not hard. We just need to factor it, i.e.: see every prime integer divisor it has in order.

20 | 2
10 | 2
5 | 5
1 | / #5*2^2#

So we know we can take a 2 out.

#sqrt(20y^5) = 2y^2sqrt(5y)#