Question

Rationalize the denominator

\[\frac{sqrt{2a}}{sqrt{a^2+1}}\]

Answer 1

\[=\frac{\sqrt{2a}}{\sqrt{a^2+1}}\]
\[=\frac{\sqrt{2a}}{\sqrt{a^2+1}}*\frac{\sqrt{a^2+1}}{\sqrt{a^2+1}}\]

Answer 2

\[\frac{\sqrt{2a(a^2+1)}}{\sqrt{a^2+1}}\]
\[\frac{\sqrt{2a^3+2a}}{a^2+1}\]

Answer 3

I don't understand the conjugate

Answer 4

What would be the conjugate of sqrt(a^2+1)?

Answer 5

I don't know

Answer 6

The conjugate is a binomial that results in an (a+b)(a-b) situation.

Answer 7

Exactly. There isn't one.

Answer 8

It's a sum of square we didn't get into that territory

Answer 9

So you multiply it by itself?

Answer 10

Just like you did because that rationalizes it. And that was the requirement of the problem. The reason you multiply by the conjugate (in appropriate circumstances) is because it rationalizes some desired quantity.

Answer 11

For this particular problem, you multiply by itself because that would get rid of the radical, but usually, you multiply by the conjugate (difference in squares) to get rid of the middle term which would contain the radical.

Answer 12

So if it was \[\sqrt{x^2-1}\] I would multiply by conjugate?

Answer 13

NO!! That does not have a conjugate.

Answer 14

\[\frac{2}{3-\sqrt{5}}\]

Answer 15

Only if it's seperate terms it has a conjugate? Okay. Thanks so much everyone!!!

Answer 16

Here's an example of when you WOULD use a conjugate:
\[\frac{\sqrt{2a}}{\sqrt{a} + \sqrt{b}}\]

Answer 17

There's a case where you would multiply by the conjugate. Do you see that it is a binomial? And the conjugate is a binomial?

Answer 18

And, my dear, please learn that there is A RAT in sepARATe

Answer 19

lol.
SepErate: Verb
SepArate: Adjective

Answer 20

Oh, lol... sorry @mertsj thanks once again!