How do you rationalize the denominator and simplify #2/(sqrt3+sqrt2)#?

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Answer 1 · 2016-03-26T14:40:43

# = 2(sqrt3 - sqrt2)#

#2 / (sqrt3 + sqrt2)#

We rationalise the expression by multiplying it with the conjugate of the denominator.

Conjugate of #color(blue)(sqrt3 + sqrt 2= sqrt 3 - sqrt2#

#(2 * color(blue)((sqrt 3 - sqrt2))) / ((sqrt3 + sqrt2) * color(blue)((sqrt 3 - sqrt2))#

# =(2 * color(blue)((sqrt 3)) + 2 * color(blue)( (- sqrt2))) / ((sqrt3)^2 - (sqrt2)^2#

# = (2sqrt3 - 2sqrt2) / ( 3-2)#

# = (2(sqrt3 - sqrt2)) / 1#

# = 2(sqrt3 - sqrt2)#