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

Question

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

1 Answer

Impact of this question

2124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-21T20:29:16

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

Explanation:

$\frac{6}{3 - \sqrt{3}}$ $6/(3-sqrt3)$
Multiplying both numerator and denominator by $3 + \sqrt{3}$ $3+sqrt3$
we get
$6 \frac{3 + \sqrt{3}}{(3 - \sqrt{3}) (3 + \sqrt{3})}$ $6(3+sqrt3)/((3-sqrt3)(3+sqrt3)$
OR
$6 \frac{3 + \sqrt{3}}{3^{2} - {(\sqrt{3})}^{2}}$ $6(3+sqrt3)/(3^2-(sqrt3)^2)$
OR
$6 \frac{3 + \sqrt{3}}{9 - 3}$ $6(3+sqrt3)/(9-3)$
OR
$6 \frac{3 + \sqrt{3}}{6}$ $6(3+sqrt3)/6$
OR
$3 + \sqrt{3}$ $3+sqrt3$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you rationalize the denominator and simplify 6/(3-sqrt3)63−√36/(3-sqrt3)?

1 Answer

Explanation:

Related questions

Impact of this question

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