How do you add #8sqrt5 - 10sqrt(1/5)#?

1 Answer
Apr 20, 2015

#8sqrt5 - 10sqrt(1/5)=#

use the property:
for any #A>0# it's true that #sqrt(1/A)=1/sqrt(A)# (in our case #A=5#)
and the fact that a fraction does not change if numerator and denominator are multiplied by the same number (in our case we multiplied them by #sqrt(5)#)

#=8sqrt5 - 10/sqrt(5)*sqrt(5)/sqrt(5)=#

obviously, #sqrt(5)*sqrt(5) = 5# in the denominator
#=8sqrt(5) - 10*sqrt(5)/5=#

reduce by #5# the second item
#=8sqrt(5)-2sqrt(5)= 6sqrt(5)#