# Can You Rationalize The Denominator 3/sqrt[6] - Sqrt[3]?

3   - √3
√6

you need to have a common denominator, so let's make √3 into a fraction of something over √6.  So let's multiply the √3 by (√6/√6).  You can always multiply something by one, since it doesn't change its value.

3    -  √3√6
√6       √6

now we can combine the numerator over the common denominator

3 - √3√6
√6

this is now a ratio.  but let's try to simply it further by multiplying top and bottom by √6

3√6 - √3√6√6
√6√6

3√6 - 6√3
6

√6 - 2√3
2

thanked the writer.
Oddman commented
You can divide out a 3 from numerator and denominator.
Anonymous commented
Oops, it got cut off. There it is.
3/sqrt[6] - sqrt[3] = (3/sqrt[6])*(sqrt[6]/sqrt[6]) - sqrt[3]
= 3sqrt[6]/6 - sqrt[3] = sqrt[6]/2 - sqrt[3]
= (sqrt[6] - 2sqrt[3])/2

thanked the writer.
3/(Sqrt[6] - Sqrt[3])

To rationalize the denominator, multiply top and bottom by the bottom's
conjugate, Sqrt[6] + Sqrt[3]. And the apply formula: (a - b) multiplied by its conjugate (a + b) is equal
to a^2 - b^2, a difference of squares.

=3(Sqrt[6] + Sqrt[3])/((Sqrt[6] - Sqrt[3])(Sqrt[6] + Sqrt[3]))
=3(Sqrt[6] + Sqrt[3])/(6-3)
=3(Sqrt[6] + Sqrt[3])/3
=Sqrt[6] + Sqrt[3]

Sqrt[6] = 2.45 and Sqrt[3]3 = 1.732/  So putting the values in the above equation you will get

=2.45 + 1.732
=4.182
thanked the writer.