How do you rationalize the dominator and simplify for the square root of 35 over the square root of 55?

1 Answer
KillerBunny
Nov 5, 2015

#sqrt(77)/11#

Explanation:

Your expression is #sqrt(35)/sqrt(55)#. Since you can always mutiply an expression by #1# without changing its value, and you can see #1# as any number divided by itself, we have

#sqrt(35)/sqrt(55)=sqrt(35)/sqrt(55)*1 = sqrt(35)/sqrt(55)*sqrt(55)/sqrt(55)#

Doing the multiplications (remember that #sqrt(a)*sqrt(b)=sqrt(a*b)#), we have

#sqrt(35*55)/sqrt(55^2)#

And of course #sqrt(a^2)=a# (if #a# is positive), so we have

#sqrt(1925)/55#.

Finally, we can factor #1925# with prime numbers, and we have

#1925 = 5^2*7*11#, and so #sqrt(1925)=sqrt(5^2*7*11)=5sqrt(7*11)#, and #5# and #55# cancel out.

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