How do you rationalize the denominator $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

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How do you rationalize the denominator $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

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Answer 1 · 2015-05-26T16:23:12

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

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

So you get:
$\frac{7 (\sqrt{3} + \sqrt{2})}{3 - 2} = 7 (\sqrt{3} + \sqrt{2})$ $(7(sqrt(3)+sqrt(2)))/(3-2)=7(sqrt(3)+sqrt(2))$

Answer 2 · 2015-05-26T16:27:08

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

$\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt(3)-sqrt(2))$

$= \frac{7 (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2})}$ $= (7(sqrt(3)+sqrt(2)))/((sqrt(3)-sqrt(2))(sqrt(3)+sqrt(2)))$

$= \frac{7 (\sqrt{3} + \sqrt{2})}{{\sqrt{3}}^{2} - {\sqrt{2}}^{2}}$ $= (7(sqrt(3)+sqrt(2)))/(sqrt(3)^2-sqrt(2)^2)$

$= \frac{7 (\sqrt{3} + \sqrt{2})}{3 - 2}$ $= (7(sqrt(3)+sqrt(2)))/(3-2)$

$= 7 (\sqrt{3} + \sqrt{2})$ $= 7(sqrt(3)+sqrt(2))$

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How do you rationalize the denominator $\frac{7}{\sqrt{3} - \sqrt{2}}$ $7/(sqrt3-sqrt2)$ ?

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