How do you rationalize the denominator and simplify #1/(sqrt3-2)#?

1 Answer
Sep 30, 2015

Multiply the numerator and denominator by bye conjugate of the denominator to get:
#color(white)("XXX")-(sqrt(3)+2)#

Explanation:

The conjugate of a two-term expression #(a+b)# is #(a-b)# and visa versa.
The product of conjugates #(a+b)xx(a-b)# is #a^2-b^2#

For the given example, the conjugate of #(sqrt(3)-2)# is #(sqrt(3)+2)#

#1/(sqrt(3)-2)#
#color(white)("XXX")=1/(sqrt(3)-2)xx(sqrt(3)+2)/(sqrt(3)+2)#

#color(white)("XXX")=(sqrt(3)+2)/((sqrt(3)^2-2^2)#

#color(white)("XXX")=(sqrt(3)+2)/(3-4)#

#color(white)("XXX")=(sqrt(3)+2)/(-1)#

#color(white)("XXX")=-(sqrt(3)+2)#