How do you rationalize the denominator and simplify #(sqrtc-sqrtd)/(sqrtc+sqrtd)#?

1 Answer
George C.
Oct 3, 2015

Multiply numerator and denominator by #(sqrt(c)-sqrt(d))# and simplify to find:

#(sqrt(c)-sqrt(d))/(sqrt(c)+sqrt(d)) = (c+d-2sqrt(cd))/(c-d)#

Explanation:

Use the difference of squares identity: #a^2-b^2 = (a-b)(a+b)# with #a = sqrt(c)# and #b=sqrt(d)#

Also use #sqrt(a)sqrt(b) = sqrt(ab)# (if #a, b >= 0#)

#(sqrt(c)-sqrt(d))/(sqrt(c)+sqrt(d))#

#= ((sqrt(c)-sqrt(d))(sqrt(c)-sqrt(d)))/((sqrt(c)-sqrt(d))(sqrt(c)+sqrt(d)))#

#= ((sqrt(c))^2-2(sqrt(c))(sqrt(d))+(sqrt(d))^2)/((sqrt(c))^2 - (sqrt(d))^2)#

#= (c+d-2sqrt(cd))/(c-d)#

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