How do you rationalize the denominator and simplify #5/(sqrt3-1)#?

2 Answers
Aug 10, 2017

To rationalise this denominator, you have to take the denominator's conjugate. The conjugate will evaluate into a difference of two squares ( #a^2 - b^2 = (a + b)(a - b)# ) which can be expressed as an integer.

So to rationalise this expression, #5/(sqrt(3)-1)#

#5/(sqrt(3)-1)#

= #5/(sqrt(3)-1) * (sqrt(3)+1)/(sqrt(3)+1)#

= #(5(sqrt(3)+1))/(3-1)#

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

Aug 10, 2017

Answer:

#color(green)((5(sqrt3+1))/2#

Explanation:

#:.5/(sqrt3-1) xx (sqrt3+1)/ (sqrt3+1)#

#sqrt3 xx sqrt3=3#

#color(white)(aaaaaaaaaaaaa)##sqrt3-1#
#color(white)(aaaaaaaaaaa)## xx underline(sqrt3+1)#
#color(white)(aaaaaaaaaaaaa)##3-sqrt3#
#color(white)(aaaaaaaaaaaaaaaa)##sqrt3-1#
#color(white)(aaaaaaaaaaaaa)##overline(3+0-1)#

#color(white)(aaaaaaaaaaaaa)##color(green)(=2#

#:.color(green)((5(sqrt3+1))/2#

~~~~~~~~~~~~~~~~~~~~~

check by calculator:

#:.5/(sqrt3-1)=color(green)(6.830127019#

#:.(5(sqrt3+1))/2=color(green)(6.830127019#