How do you rationalize the denominator and simplify #2/(5-sqrt3)#?

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Answer 1 · 2018-05-18T13:37:50

See a solution process below:

To rationalize the denominator we must multiply the fraction by the appropriate form of #1# to eliminate the radicals from the denominator.

For a denominator of the type #(a color(red)(-) b)# we need to multiply by #(a color(red)(+) b)#:

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

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

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

#(10 + 2sqrt(3))/(25 - 0 - 3) =>#

#(10 + 2sqrt(3))/22 =>#

#(5 + sqrt(3))/11 =>#

Or

#5/11 + sqrt(3)/11#

Answer 2 · 2018-05-18T13:51:15

#" "#
#color(brown)(2/(5-sqrt3) = [5+sqrt(3)]/11#

#" "#
Rationalize the denominator and simplify: #color(blue)(2/(5-sqrt3)#

#"Given the rational expression: " color(red)(2/(5-sqrt3)#

#rArr 2/(5-sqrt3)*color(blue)([(5+sqrt3)/(5+sqrt3)]#

#rArr [(2*(5+sqrt(3)))/((5^2)-(sqrt(3))^2]]#

Identity used: :#color(green)((a+b)(a-b)=a^2-b^2#

In our problem: #a=5 and b= sqrt(3)#

#rArr (2*(5+sqrt(3)))/(25-3)#

#rArr (2(5+sqrt(3)))/(22)#

#rArr (cancel 2(5+sqrt(3)))/(cancel 22^color(blue)(11))#

#rArr (5+sqrt(3))/11#

Hence,

#color(brown)(2/(5-sqrt3) = [5+sqrt(3)]/11#

Hope it helps.