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

1 Answer
Jun 12, 2015

Answer:

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

Explanation:

#(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)#