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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. :lol: Thanks.
6/sqr(x) + sqr(3)
6sqr(x)/x + sqr(3)
You posted:
Alkatran answered your first question.Quote:
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.
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.
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.
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. :lol:
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.