How do you rationalize the denominator and simplify #(x-3)/(sqrtx-sqrt3)#?

2 Answers
Sep 26, 2016

Answer:

To rationalize a denominator in the form of #sqrta - sqrtb#, you multiply the fraction by 1 in the form #(sqrta + sqrtb)/(sqrta + sqrtb)#

Explanation:

The reason for doing this practice comes from general form for factoring binomials that contain the difference two squares:

#a^2 - b^2 = (a - b)(a + b)#

Returning to the given fraction, we multiply by 1 in form #(sqrtx + sqrt3)/(sqrtx + sqrt3)#

#(x - 3)/(sqrtx - sqrt3)(sqrtx + sqrt3)/(sqrtx + sqrt3) = #

#((x - 3)(sqrtx + sqrt3))/(x - 3) = #

#sqrtx + sqrt3#

Sep 26, 2016

Answer:

#sqrt x + sqrt 3#

Explanation:

divide the Numerator and denominator by #sqrtx + sqrt 3#.
we get, # ( x - 3)/(sqrt x - sqrt 3) * (sqrt x + sqrt 3)/(sqrt x + sqrt 3)#
= #[(x - 3)(sqrt x + sqrt 3)]/[(sqrt x)^2 - (sqrt 3)^2] = [(x - 3)(sqrt x + sqrt 3)]/(x - 3) = sqrt x + sqrt 3#