How do you rationalize the denominator and simplify #(x-3)/(sqrtx-sqrt3)#?
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To rationalize a denominator in the form of #sqrta - sqrtb#, you multiply the fraction by 1 in the form #(sqrta + sqrtb)/(sqrta + sqrtb)#
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#
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#
