How do you rationalize the denominator and simplify #(sqrt 6 - 3 ) / 4#?
1 Answer
May 4, 2016
denominator is already rationalised
Explanation:
The set of rational numbers Q are expressed in the form
#a/b# where a,b
#inZ, b≠0# now 4 is in this form ie.
#4/1rArr " in rational form "# Normally require to rationalise the denominator when the radical is on the denominator, which is not the case here.
For example , if
#sqrt6 -3 " was on the denominator "# Then to rationalise , we multiply by it's Conjugate
#• " conjugate of "sqrta ± b " is " sqrta ∓ b# and multiplying by the conjugate produces a rational number
#rArr(sqrt6-3)(sqrt6+3)=(sqrt6)^2+3sqrt6-3sqrt6-9#
#=6-9=-3" a rational value "#