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

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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 $sqrt(a^2)=absa$ .

$(7sqrt53)/53$

How do you rationalize the denominator and simplify sqrt49/sqrt53sqrt49/sqrt53?