How do you rationalize the denominator and simplify #sqrt49/sqrt53#?

Question

1310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-30T07:06:24

The answer is #(7sqrt53)/53# .

#sqrt49/sqrt53#

Rationalize the denominator.

#(sqrt49sqrt53)/(sqrt53sqrt53)#

Apply the rule #sqrtasqrta=a# .

#(sqrt49sqrt53)/(53)#

Write the prime factors for #sqrt49# and #sqrt53#.

#(sqrt(7xx7)sqrt(53)# (#53# is a prime number)

Rewrite #7xx7# as #7^2#.

#(sqrt(7^2)sqrt53)/53=#

Apply the rule #sqrt(a^2)=absa# .

#(7sqrt53)/53#