How do you simplify #(sqrt2 /sqrt8) div (sqrt2 /sqrt40)#?

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Answer 1 · 2018-05-14T15:27:06

#sqrt(5)#

Dividing by a fraction is the same as multiplying by the inverse of the fraction:

#a/b : c/d = a/b \cdot d/c#

So, in your case,

#sqrt(2)/sqrt(8) : sqrt(2)/sqrt(40) = sqrt(2)/sqrt(8) \cdot sqrt(40)/sqrt(2)#

We can cross-cancel #sqrt(2)#:

#cancel(sqrt(2))/sqrt(8) \cdot sqrt(40)/cancel(sqrt(2)) = sqrt(40)/sqrt(8)#

Finally, remembering that #sqrt(ab)=sqrt(a)sqrt(b)#, we have

#sqrt(40)/sqrt(8) = sqrt(5 \cdot 8)/sqrt(8) = (sqrt(5)\cdot cancel(sqrt(8)))/cancel(sqrt(8)) = sqrt(5)#