Help!! http://prntscr.com/1c95zm
\[y + 1/y = 4\]Solve that for \(y\) by multiplying both sides by \(y\) \[y^2 + 1 = 4y\]Now you've got a quadratic which you should be able to solve. Take the answers you get and plug them into the other expression and see what you get.
or, why don't you write y^2 + 1/y^2 as y^2 + 1/y^2 + 2 - 2 see if this helps ?
@shubhamsrg issay se kya karun? :@
(y^2) + (1/y)^2 + 2(y)(1/y) - 2 you see an identity hidden somewhere over there? :O
no i am not getting this :@
You understand how I expressed your given equation in this form ? I mean everything till now is clear ?
no :P
y^2 + 1/y^2 = (y)^2 + (1/y)^2 = (y)^2 + (1/y)^2 + 2 -2 . You must be comfortable till here ?
yeah!
I wrote 2 as 2(y)(1/y)
which brings us to (y)^2 + (1/y)^2 + 2(y) (1/y) -2
yeah...
can you tell me expansion of (y + 1/y)^2 ?
y2+2+1/y2
yes, and what have we written above ?
oh ok now i get this!
glad you do (Y)
so what's the answer? prove that you get it :-)
perhaps we took a longer road, what you could have easily understood was, squaring both sides in (y+1/y) =4
-.- why didn't you give me this easy explanation before?
hmm, both things are same.
what? should I apologize? :P
nah! i'll spare you :) but could you continue your longer explanation, i mean how would it lead to the same answer?
(y+1/y)^2= 4^2 y^2 + 1/y^2 + 2 = 16 so y^2 + 1/y^2 = ...
and thanks for sparing me .-.
(y)^2 + (1/y)^2 + 2 -2 ye wali eq kaisay 16 answer de gi?
nahi, dono 14 denge! o.O
kaisayy??
(y + 1/y)^2 -2 first case me =14 2nd case me to 14 hai hi clearly
ah ok i am sorry i got this now :P
(Y)
they threw you a little curve ball with the answer choices :-)
yeah !
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