How do you rationalize the denominator $7/(sqrt3-sqrt2)$ ?

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Answer 1 · 2015-05-26T16:23:12

You can multiply and divide your expression by $sqrt(3)+sqrt(2)$ to get:
$7/(sqrt(3)-sqrt(2))*(sqrt(3)+sqrt(2))/(sqrt(3)+sqrt(2))=$

in the denominator you have a notable:
$(a+b)(a-b)=a^2-b^2$

So you get:
$(7(sqrt(3)+sqrt(2)))/(3-2)=7(sqrt(3)+sqrt(2))$

Answer 2 · 2015-05-26T16:27:08

Multiply numerator and denominator by the conjugate $(sqrt(3)+sqrt(2))$ :

$7/(sqrt(3)-sqrt(2))$

$= (7(sqrt(3)+sqrt(2)))/((sqrt(3)-sqrt(2))(sqrt(3)+sqrt(2)))$

$= (7(sqrt(3)+sqrt(2)))/(sqrt(3)^2-sqrt(2)^2)$

$= (7(sqrt(3)+sqrt(2)))/(3-2)$

$= 7(sqrt(3)+sqrt(2))$

How do you rationalize the denominator 7/(sqrt3-sqrt2) 7/(sqrt3-sqrt2) ?