How do you rationalize the denominator and simplify #sqrt 55/sqrt10#?

1 Answer
Apr 22, 2015

To keep the numbers smaller, try to reduce first. Both #55# and #10# are divisible by #5#. Use that to write:

#sqrt55/sqrt10 = (sqrt5sqrt11)/(sqrt5sqrt2)= sqrt11/sqrt2#

Now rationalize tlhe denominator by multiplying by #1# in the form: #sqrt2/sqrt2#

#sqrt55/sqrt10 = sqrt11/sqrt2 sqrt2/sqrt2 = sqrt22/2# and we're done.

Notice

I find it worth trying to reduce first. If you don't reduce first, then you'll still get the correct answer, and it hooks like this:

#sqrt55/sqrt10 =sqrt55/sqrt10 sqrt10/sqrt10 = sqrt550/10#

#sqrt55/sqrt10 = sqrt550/10 = sqrt(55*10)/10 = sqrt(5*11*2*5)/10 = sqrt (25*22)/10 = (5sqrt22)/10 = sqrt22/2#