How do you rationalize the denominator and simplify #2/(2sqrt3-4)#?

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Answer 1 · 2018-05-01T23:44:12

#-sqrt3+2#

#2/(2sqrt3-4)#

First, factor #2# from the denominator:
#2/(2(sqrt3-2))#

Both numerator and denominator have a #2#, so we can cancel them out:
#1/(sqrt3-2)#

To rationalize the denominator, we multiply both numerator and denominator it by its conjugate, or #sqrt3+2#.

Let's do that:
#1/(sqrt3 - 2) color(blue)(xx(sqrt3+2)/(sqrt3+2))#

Simplify:
#(sqrt3-2)/(sqrt3^2 - 2sqrt3 + 2sqrt3 - 2^2)#

#(sqrt3-2)/(3-4)#

#(sqrt3-2)/(-1)#

#-(sqrt3-2)#

So the final answer is:
#-sqrt3+2#

Hope this helps!