How do you simplify #7/(5+sqrt3)#?

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Answer 1 · 2017-03-31T09:22:29

#=35/22-(7sqrt3)/22#

You must know that the conjugate of an irrational number of the type
#a+sqrtb# is given by #a-sqrtb# and that of #a-sqrtb# is given by #a+sqrtb#.

#7/(5+sqrt3)#

(multiplying and dividing by conjugate of #5+sqrt3#)

#= 7/(5+sqrt3)*(5-sqrt3)/(5-sqrt3)#

#= (7(5-sqrt3))/(5^2-(sqrt3)^2)#

#=(7(5-sqrt3))/22#
#=35/22-(7sqrt3)/22#

Answer 2 · 2017-03-31T09:33:42

Rationalise the denominator by multiplying by the conjugate surd.

#7/(5+sqrt3)# = #7/(5 + sqrt3) * (5-sqrt3)/(5-sqrt3)#

= #[7(5-sqrt3)]/[(5^2)-(sqrt3)^2]#

= #(35 - 7sqrt3)/(25 - 3)#

= #(35 - 7sqrt3)/22#