How do you rationalize $3 / (sqrt5 -sqrt2)$ ?

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Answer 1 · 2015-05-14T19:11:16

You can rationalize the denominator by multiplying top and bottom by the conjugate $sqrt(5) + sqrt(2)$ as follows:

$3/(sqrt(5)-sqrt(2))$

$= (3*(sqrt(5)+sqrt(2)))/((sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2))$

$= (3*(sqrt(5)+sqrt(2)))/(sqrt(5)sqrt(5)-sqrt(2)sqrt(2))$

$= (3*(sqrt(5)+sqrt(2)))/(5-2)$

$= (3*(sqrt(5)+sqrt(2)))/3$

$= sqrt(5)+sqrt(2)$

How do you rationalize 3 / (sqrt5 -sqrt2) 3 / (sqrt5 -sqrt2) ?