How do you rationalize the denominator and simplify $\frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}}$ $(6sqrt3 - sqrt7) /( 4 - sqrt3)$ ?

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How do you rationalize the denominator and simplify $\frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}}$ $(6sqrt3 - sqrt7) /( 4 - sqrt3)$ ?

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Answer 1 · 2017-11-19T17:08:26

$\frac{(4 + \sqrt{3}) (6 \sqrt{3} - \sqrt{7})}{13}$ $((4+sqrt3)(6sqrt3-sqrt7))/13$

Explanation:

$\frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}} = \frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}} (\frac{4 + \sqrt{3}}{4 + \sqrt{3}})$ $(6sqrt(3)-sqrt(7))/(4-sqrt(3))=(6sqrt(3)-sqrt(7))/(4-sqrt(3))((4+sqrt3)/(4+sqrt3))$

$= \frac{(4 + \sqrt{3}) (6 \sqrt{3} - \sqrt{7})}{16 - 3}$ $=((4+sqrt3)(6sqrt3-sqrt7))/(16-3)$

$= \frac{(4 + \sqrt{3}) (6 \sqrt{3} - \sqrt{7})}{13}$ $=((4+sqrt3)(6sqrt3-sqrt7))/13$

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How do you rationalize the denominator and simplify $\frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}}$ $(6sqrt3 - sqrt7) /( 4 - sqrt3)$ ?

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How do you rationalize the denominator and simplify $\frac{6 \sqrt{3} - \sqrt{7}}{4 - \sqrt{3}}$ $(6sqrt3 - sqrt7) /( 4 - sqrt3)$ ?