How do you rationalize the denominator and simplify $\frac{2}{2 \sqrt{3} - 4}$ $2/(2sqrt3-4)$ ?

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Answer 1 · 2018-05-01T23:44:12

$- \sqrt{3} + 2$ $-sqrt3+2$

$\frac{2}{2 \sqrt{3} - 4}$ $2/(2sqrt3-4)$

First, factor $2$ $2$ from the denominator:
$\frac{2}{2 (\sqrt{3} - 2)}$ $2/(2(sqrt3-2))$

Both numerator and denominator have a $2$ $2$ , so we can cancel them out:
$\frac{1}{\sqrt{3} - 2}$ $1/(sqrt3-2)$

To rationalize the denominator, we multiply both numerator and denominator it by its conjugate, or $\sqrt{3} + 2$ $sqrt3+2$ .

Let's do that:
$\frac{1}{\sqrt{3} - 2} \times \frac{\sqrt{3} + 2}{\sqrt{3} + 2}$ $1/(sqrt3 - 2) color(blue)(xx(sqrt3+2)/(sqrt3+2))$

Simplify:
$\frac{\sqrt{3} - 2}{{\sqrt{3}}^{2} - 2 \sqrt{3} + 2 \sqrt{3} - 2^{2}}$ $(sqrt3-2)/(sqrt3^2 - 2sqrt3 + 2sqrt3 - 2^2)$

$\frac{\sqrt{3} - 2}{3 - 4}$ $(sqrt3-2)/(3-4)$

$\frac{\sqrt{3} - 2}{- 1}$ $(sqrt3-2)/(-1)$

$- (\sqrt{3} - 2)$ $-(sqrt3-2)$

So the final answer is:
$- \sqrt{3} + 2$ $-sqrt3+2$

Hope this helps!

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