How do you rationalize $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ $(sqrt3-sqrt2)/ ( sqrt3+sqrt2)$ ?

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How do you rationalize $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ $(sqrt3-sqrt2)/ ( sqrt3+sqrt2)$ ?

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Answer 1 · 2015-05-16T19:51:11

You can multiply and divide by: $\sqrt{3} - \sqrt{2}$ $sqrt(3)-sqrt(2)$ to get:
$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} =$ $(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2))*(sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))=$
$= \frac{{(\sqrt{3} - \sqrt{2})}^{2}}{3 - 2} = 3 - 2 \sqrt{6} + 2 = 5 - 2 \sqrt{6}$ $=(sqrt(3)-sqrt(2))^2/(3-2)=3-2sqrt(6)+2=5-2sqrt(6)$

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How do you rationalize $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ $(sqrt3-sqrt2)/ ( sqrt3+sqrt2)$ ?

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How do you rationalize $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ $(sqrt3-sqrt2)/ ( sqrt3+sqrt2)$ ?