How do you rationalize the denominator $sqrt6/(sqrt13+3)$ ?

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Answer 1 · 2015-06-02T20:06:23

Multiply both numerator (top) and denominator (bottom) by the conjugate $(sqrt(13)-3)$ ...

$sqrt(6)/(sqrt(13)+3)$

$=sqrt(6)/(sqrt(13)+3)*(sqrt(13)-3)/(sqrt(13)-3)$

$=(sqrt(6)*(sqrt(13)-3))/((sqrt(13)+3)(sqrt(13)-3))$

$=(sqrt(6)*(sqrt(13)-3))/(13-9)$

$=(sqrt(6)*(sqrt(13)-3))/4$

How do you rationalize the denominator sqrt6/(sqrt13+3)sqrt6/(sqrt13+3)?