How do i rationalize this denominator?

[Dillinger4]

How do i rationalize the denominator in this? 6/sqr(x) + sqr(3) I know for somthing like 7/sqr(4) i can use a rationalizing factor of sqr(4) to end up getting 7sqr(4)/4 but then im not sure if this ends up being reduced to 7sqr(1). So i guess im asking two quesitons. Thanks.

[alkatran]

6/sqr(x) + sqr(3)
6sqr(x)/x + sqr(3)

[RAEsquivelC]

You posted:

How do i rationalize the denominator in this? 6/sqr(x) + sqr(3) I know for somthing like 7/sqr(4) i can use a rationalizing factor of sqr(4) to end up getting 7sqr(4)/4 but then im not sure if this ends up being reduced to 7sqr(1). So i guess im asking two quesitons. Thanks.

Alkatran answered your first question.

As to your second question, yes, you do "end up getting 7sqr(4)/4, but, that cannot be reduced to 7sqr(1), as sqr(4)/4 is NOT equal to sqr(1). Sqr(4)/4 has two answers, 1/2 and -1/2, so, 7/sqr(4), or, 7*sqr(4)/4, equals +7/2 and -7/2.

[Dillinger4]

alkatran why do we only use sqr(x) as the rationalizing factor? 6sqr(x)/x + sqr(3) still would not be in proper form because of the radical in the denominator.

[Dillinger4]

So with 6sqr(4)/2 you can either find the square of the numerator then divide by the denom and reduce. ie 6sqr(4)/2 --> 12/2 --> 6
or you can just reduce from the start 6sqr(4)/2 --> 3sqr(4) --> 6. Sounds right i guess.

[Dillinger4]

alkatran for 6/(sqr(x) + sqr(3)) it seems that i would have to multiply the denominator by it's conjugate ie.(sqr(x) - sqr(3)) which would give me the difference of two squares. Then do the same with the numerator. 6/sqr(x) + sqr(3) --> 6(sqr(x)-sqr(3))/(sqr(x) + sqr(3))(sqr(x)-sqr(3))
--> 6(sqr(x)-sqr(3))/x-3.