rationalizing numerator?

Question

how do i rationalize the numerator if the problem is as such (

3. (sqroot 6 - 3 ) / 4
4. (sqroot 5 + sqroot 6 ) / 3

thanks, much appreciation for the help on the other post i had. seem to understand it a little better now. but for the numerator i thought that if you already had the square root sign on top of the fraction then its already rationalize or isn't it?.

thanks again..

i thought that you can't have square roots on the bottom of the fraction?

Answer 1

You can use the same technique for rationalizing the denominator.

Answer 2

Remember that (a+b)*(a-b) = a^2 - b^2 1) Multiply numerator and denominator by [sqrt(5) - sqrt(6)]. The denominator becomes the 5 - 6, or -1, so the rationalized result is -3*[sqrt(5) - sqrt(6)] 2) same thing, except multiply by [sqrt(14) + 2] to get 14-4 in the denominator Rationalize the numerators in 3 and 4 the same way: the numerators will come out -3 and -1 respectively

Answer 3

In general, a rationalized fraction has no root signs in the denominator. However, there is nothing "magic" about a rationalized fraction. and if you are asked to rationalize the numerator, i take that to mean to get rid of root signs in the numerator. In most cases, you can rationalize both numerator and denominator, so when you do the numerator, you can expect root signs in the denominator.