How do you rationalize the denominator and simplify #4/(sqrt7-sqrt5)#?

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Answer 1 · 2017-11-12T18:01:08

#=2(sqrt7+sqrt5)#

#4/(sqrt7-sqrt5)#

to rationalise the denominator we make use of

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

so

#4/(sqrt7-sqrt5)= 4/(sqrt7-sqrt5)xx (sqrt7+sqrt5)/(sqrt7+sqrt5)#

#=(4(sqrt7+sqrt5))/((sqrt7)^2-(sqrt5)^2)#

#=(4(sqrt7+sqrt5))/(7-5)#

#=(cancel(4)^2(sqrt7+sqrt5))/cancel(2)#

#=2(sqrt7+sqrt5)#