How do you evaluate #(5- 2sqrt ( 3))/( 3 + sqrt ( 3))#?

1 Answer
George C.
Sep 28, 2017

#(5-2sqrt(3))/(3+sqrt(3)) = 7/2-11/6sqrt(3)#

Explanation:

I think you might have intended:

#(5-2sqrt(3))/(3+sqrt(3))#

If so, then we can rationalise the denominator by multiplying both numerator and denominator by the radical conjugate #3-sqrt(3)# as follows:

#(5-2sqrt(3))/(3+sqrt(3)) = ((5-2sqrt(3))(3-sqrt(3)))/((3+sqrt(3))(3-sqrt(3)))#

#color(white)((5-2sqrt(3))/(3+sqrt(3))) = ((5)(3)+(5)(-sqrt(3))+(-2sqrt(3))(3)+(-2sqrt(3))(-sqrt(3)))/(3^2-(sqrt(3))^2)#

#color(white)((5-2sqrt(3))/(3+sqrt(3))) = (15-5sqrt(3)-6sqrt(3)+6)/(9-3)#

#color(white)((5-2sqrt(3))/(3+sqrt(3))) = (21-11sqrt(3))/6#

#color(white)((5-2sqrt(3))/(3+sqrt(3))) = 7/2-11/6sqrt(3)#

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