How do you simplify #((sqrt7) - (sqrt 2))/((sqrt7) + (sqrt 2))#?

1 Answer
Apr 9, 2015

For this kind of a question you should multiply the nominator and the denominator with the demonator but with the minus sign. (forgive me for my bad english :). What i mean is, this is what you should do;

#((sqrt7)-(sqrt2))/((sqrt7)+(sqrt2))# since there is #((sqrt7)+(sqrt2))# in the denomintor you should multiply the whole equation with #((sqrt7)-(sqrt2))#

The reason we do this to reach this result in the denominator part;
#((sqrt7)+(sqrt2))# . #((sqrt7)-(sqrt2))# = # 7 - 2 # = #5#
I came to that from this formula;
#(x+y) . (x-y) = x^2 - y^2 # # => # This is a formula that should be memorized in order to solve this kind of question.

Also you have to know this too;
#(x-y)^2 = x^2 - 2xy + y^2 #

So if i solve the entire question the result will be;
#((sqrt7)-(sqrt2))/((sqrt7)+(sqrt2))# = #((sqrt7)-(sqrt2))/((sqrt7)+(sqrt2)) * ((sqrt7)-(sqrt2))/((sqrt7)-(sqrt2))# = #((sqrt7)-(sqrt2))^2/(7-2)# = #(7-2sqrt14 + 2)/ 5# = #(9- 2sqrt14)/5# #=># this is the simplest version i can write. I hope it helps.