How do you rationalize the denominator #(2+sqrt3)/(2-sqrt3)#?

1 Answer
Alan P.
May 29, 2015

The conjugate of a binomial #(a+b)# is #(a-b)#;
note that the product of such conjugates #(a+b)(a-b) = a^2-b^2#

and this allows us to clear radicals (for example from a denominator; although this requires that the radical show up some where else; in this case in the numerator).

Using the given #(2+sqrt(3))/(2-sqrt(3))# for demonstration purposes

If we multiply the denominator by the conjugate of the denominator (we will also have to multiply the numerator by the same amount to avoid changing the value of the expression), we get

#(2+sqrt(3))/(2-sqrt(3))xx(2+sqrt(3))/(2+sqrt(3)#

#= ((2+sqrt(3))xx(2+sqrt(3)))/(2^2-(sqrt(3)^2)#

#= (4 +2sqrt(3) +3)/(4-3)#

#=7+2sqrt(3)#

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