How do you simplify #sqrt2 + sqrt( 2/49)#?

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Answer 1 · 2018-05-09T23:28:24

#(8sqrt2)/7#

#sqrt2# has no perfect squares in it, so we can leave it the way it is for now. But we can rewrite this entire expression as

#sqrt(2)+(sqrt2/sqrt49)#

(because of the property #sqrt(a/b)=sqrta/sqrtb#)

Which simplifies to

#sqrt2+sqrt2/7#

We can factor out a #sqrt2# to get

#sqrt2(1+1/7)#

#=sqrt2(1 1/7)#

Changing to an improper fraction to get

#=sqrt2(8/7)#

which finally simplifies to

#(8sqrt2)/7#

Hope this helps!