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

2 Answers

Answer:

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

Explanation:

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#

Mar 31, 2017

Answer:

Rationalise the denominator by multiplying by the conjugate surd.

Explanation:

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

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

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

= #(35 - 7sqrt3)/22#