How do you rationalize the denominator and simplify #(4-sqrt3)/(6+sqrty)#?

1 Answer
Feb 12, 2017

Answer:

#(4-sqrt3)/(6+sqrty)=(24-4sqrty-6sqrt3+sqrt(3y))/(36-y)#

Explanation:

To rationalize the denominator, we multiply numerator and denominator by the conjugate of its denominator .

Conjugate of an irrational number like #sqrta+sqrtb# is #sqrta-sqrtb# and that of #p+sqrtq# is #p-sqrtq#.

Here in #(4-sqrt3)/(6+sqrty)#, denominator is #6+sqrty# and assuming #y# is rational positive number, conjugate of denominator is #6-sqrty#. As such

#(4-sqrt3)/(6+sqrty)#

= #(4-sqrt3)/(6+sqrty)xx(6-sqrty)/(6-sqrty)#

= #((4-sqrt3)(6-sqrty))/(6^2-(sqrty)^2)#

= #(4xx6-4sqrty-6sqrt3+sqrt(3y))/(6^2-(sqrty)^2)#

= #(24-4sqrty-6sqrt3+sqrt(3y))/(36-y)#

Note - In case you have just #sqrtp# in denominator, just multiply numerator and denominator by #sqrtp#.