What is #(sqrt7- 5)/(sqrt2 + 5 )#?

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Answer 1 · 2015-06-12T19:21:29

The answer is #-(sqrt14-5sqrt7-5sqrt2+25)/(23)#.

#(sqrt 7-5)/(sqrt 2+5)#

Rationalize the denominator by multiplying the numerator and denominator by #(sqrt2-5)#.

#(sqrt 7-5)/(sqrt 2+5)*(sqrt 2-5)/(sqrt 2-5)# =

#((sqrt 7-5)(sqrt 2-5))/((sqrt 2+5)(sqrt 2-5))# =

The denominator is in the form of the difference of squares: #a^2-b^2#

#((sqrt 7-5)(sqrt 2-5))/((sqrt2)^2-5^2)# =

#((sqrt 7-5)(sqrt 2-5))/(2-25)# =

#((sqrt 7-5)(sqrt 2-5))/(-23)#

FOIL the numerator.

#(sqrt14-5sqrt7-5sqrt2+25)/(-23)# =

#-(sqrt14-5sqrt7-5sqrt2+25)/(23)#