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

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

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Answer 1 · 2018-03-02T01:57:08

$\frac{\sqrt{21} + 2 \sqrt{3}}{3}$ $(sqrt21+2sqrt3)/3$

Explanation:

To rationalize the denominator, the best move is to multiply the numerator and denominator by the reciprocal of the denominator (which is the denominator but with the middle sign reversed).

$\frac{\sqrt{3}}{\sqrt{7} - 2} \cdot \frac{\sqrt{7} + 2}{\sqrt{7} + 2}$ $sqrt3/(sqrt7-2) * color(blue)((sqrt7+2)/(sqrt7+2))$

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

From here, we can simplify the numerator and denominator.

$\sqrt{α} \sqrt{β} = \sqrt{α β} \to \sqrt{3} \sqrt{7} = \sqrt{21}$ $sqrt(alpha)sqrt(beta)=sqrt(alphabeta) -> sqrt3sqrt7=sqrt21$

$\sqrt{3} (\sqrt{7} + 2) = \sqrt{21} + 2 \sqrt{3}$ $sqrt3(sqrt7+2)=color(red)(sqrt21+2sqrt3)$

$(sqrt7-2)(sqrt7+2) = 7+cancel(2sqrt7-2sqrt7)-4 = color(red)3$

$sqrt3/(sqrt7-2)=color(red)((sqrt21+2sqrt3)/3)$

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

1 Answer

Explanation:

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How do you rationalize the denominator and simplify sqrt(3)/(sqrt(7)-2)√3√7−2sqrt(3)/(sqrt(7)-2)?

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Explanation:

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