How do you rationalize the denominator of #(6x^4)/(sqrt(7x-1))#?

1 Answer
Jan 26, 2017

#((6x^4)(sqrt(7x-1)))/(7x-1)#

Explanation:

To rationalize a denominator, we multiply, top and bottom, by the same value as the denomintor. That makes it #sqrt(7x+1)* sqrt(7x-1)# or #(sqrt(7x-1))^2#, which is also #7x-1#. In this case, our problem works the same way:

#(6x^4)/sqrt(7x-1)#*#sqrt(7x-1)/(sqrt(7x-1))#=#((6x^4)sqrt(7x-1))/(sqrt(7x-1)0^2#=#((6x^4)sqrt(7x-1))/(7x-1)#