How do you simplify #sqrt(2 / 75)#?

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Answer 1 · 2018-03-30T00:37:18

#=>1/5sqrt(2/3)#

#=>sqrt(2/75)#

#=>sqrt(2/(3*25)#

#=>sqrt(2/(3*5^2)#

#=>1/5sqrt(2/3)#

Answer 2 · 2018-03-30T00:43:45

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1) Rewrite problem
2)Take the biggest perfect square out of the bottom square root
3) Rationalize denominator

Answer 3 · 2018-03-30T13:55:51

#sqrt 6/15#

#sqrt(2/75)#

#:.=sqrt2/sqrt 75xxsqrt 75/sqrt 75#

#:.=sqrt(2*75)/75#

#:.=sqrt150/75#

#:.=sqrt (2*3*5*5)/75#

#:.sqrt5*sqrt5=5#

#:.=(cancel5^1 sqrt 6)/cancel75^15#

#:.=sqrt6/15#