What is the simplest radical form of #sqrt160#?

2 Answers
Aug 5, 2016

Answer:

#4sqrt10#

Explanation:

Write 160 as the product of its prime factors, then we know what we are dealing with.

#sqrt160 = sqrt(2xx2xx2xx2xx2xx2xx5) = sqrt(2^5 xx 5)#

=#sqrt(2^5 xx 5) = sqrt(2^4 xx 2 xx 5)#

=#4sqrt10#

Aug 5, 2016

Radicals can be split by multiplication. It helps to be able to find perfect squares underneath the radicals during the factorization, and #16# is a convenient perfect square.

If it helps, try going in steps of factoring out #2#.

#sqrt(160)#

#sqrt(2*80)#

#sqrt(2*2*40)#

#sqrt(2*2*2*20)#

#sqrt(2*2*2*2*10)#

#= sqrt(16*10)#

#= sqrt(16)*sqrt(10)#

Since #sqrt(16) = 4#, we end up with #color(blue)(4sqrt10)#.