How do you rationalize the denominator $\frac{\sqrt{6}}{\sqrt{13} + 3}$ $sqrt6/(sqrt13+3)$ ?

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Answer 1 · 2015-06-02T20:06:23

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

$\frac{\sqrt{6}}{\sqrt{13} + 3}$ $sqrt(6)/(sqrt(13)+3)$

$= \frac{\sqrt{6}}{\sqrt{13} + 3} \cdot \frac{\sqrt{13} - 3}{\sqrt{13} - 3}$ $=sqrt(6)/(sqrt(13)+3)*(sqrt(13)-3)/(sqrt(13)-3)$

$= \frac{\sqrt{6} \cdot (\sqrt{13} - 3)}{(\sqrt{13} + 3) (\sqrt{13} - 3)}$ $=(sqrt(6)*(sqrt(13)-3))/((sqrt(13)+3)(sqrt(13)-3))$

$= \frac{\sqrt{6} \cdot (\sqrt{13} - 3)}{13 - 9}$ $=(sqrt(6)*(sqrt(13)-3))/(13-9)$

$= \frac{\sqrt{6} \cdot (\sqrt{13} - 3)}{4}$ $=(sqrt(6)*(sqrt(13)-3))/4$

How do you rationalize the denominator √6√13+3sqrt6/(sqrt13+3)?