How do you rationalize the denominator and simplify #(sqrt 6 - 3 ) / 4#?

1 Answer
May 4, 2016

Answer:

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 "#