How do you rationalise the denominator of #(5+sqrt(6))/(5-sqrt(6))# ?
1 Answer
Feb 2, 2018
Explanation:
Note that:
#a^2-b^2 = (a-b)(a+b)#
Hence we can rationalise the denominator by multiplying both numerator and denominator by the radical conjugate
#(5+sqrt(6))/(5-sqrt(6)) = ((5+sqrt(6))(5+sqrt(6)))/((5-sqrt(6))(5+sqrt(6)))#
#color(white)((5+sqrt(6))/(5-sqrt(6))) = (5^2+2(5)sqrt(6)+(sqrt(6))^2)/(5^2-(sqrt(6))^2)#
#color(white)((5+sqrt(6))/(5-sqrt(6))) = (25+10sqrt(6)+6)/(25-6)#
#color(white)((5+sqrt(6))/(5-sqrt(6))) = (31+10sqrt(6))/19#
#color(white)((5+sqrt(6))/(5-sqrt(6))) = 31/19+10/19 sqrt(6)#
