1let y= √7+√7-................∞
therefore y=√7+y
y2 = 7 + y
so, y=1+or_√1-4.1.-7 divided by 2
this will give u y=3 or 2
23 Answers
62dont worry about all that crap ....
I was only talking about problems like
x4+x3+x2+x+1 = 0
try getting the roots of this one!
62sindhu there is a small mistake here!!
it is not all +++++ look carefully it is + - + - + -
so the sum you have found is for √7+√7+√7+√7+√7+√7+√7..............
Not for the quesiton given!!
162you can but.. the problem here is that you will go back to the initial queston..
It will be different but solvable
62hey akand.. i only meant the biquadratic.. not azinkya's problem!!
1u mean this can be an IIT ques???????????? coooooolll......
I kno it may not be tht simple.... itll be a small part of a big ques... i guess hehe
62good work akand :)
this is the point i was trying to say..
see not all questions can be solved..
but some like these.. can definitely come in IIT JEE!! (You have to be very observant!)
1wel v can solve tht eqn by dividing by x2...
tht will becum..x2+x+1+1/x+1/x2=0
then (x2+1/x2)+(x+1/x)+1=0
then take x+1/x as t then square t... ull get d answer.......
62No Ajinkya..
you cant solve it like this.. whn you use the four equations again, you will get back the same biquadratic equation.... So all the hardwork will biring you back to square 1
I gave this problem only to demonstrate that there is a very narrow band of problems that can be solved very easily. You dont need any JEE knowledge!
1i just know ax4 +bx3 +cx2 + dx +e =0
tat consider eqn
(x-α)(x-β)(x-γ)(x-δ) = 0
ten roots may b 4 or 3 or 2 or 1
i mean 1 or 2 or 3 or all can b same
then αβγδ = e/a
Σαβγ = - d/a
Σαβ = c/a
Σα = -b/a
using this v can get but it will b a bigggggg quadra an simultaneous
lrngggggggggggggggghhhhty metod
11Do you mean synthtic division? That's one heck of a lot to explain in posts and replies!
62no dude.. forth powers cannot be solved! in general
Even 3rd powers are not in syllabus(there is a standard way to solve it though!)
The only way you can solve a biquadratic at this level is if you divide by x2
If you get somethng good it will serve.. otherwise that is not in syllabus
1lodik nice tanx nishant an anirudh
u guys ........
u solved the whole eqn of power 4 ,2 ,1 ,0
then u can find ...............>>
but how can u find three rots . it was a realllly humbag
i thing i ask anirudh is ok if u had not seeen 3 then it would had ben a lot of dffcult . and it would take much of time . any shotkutz>>??
62no
actually see
i tried (7-x) as perfect square
one is x=3 and another is x=-2
!!
so u see where that comes from!
bcos for one case we take the +ve root ie root of 4
but in the other case ..(only my guess!) root will be of (7--2) = 9 = -3!!!
All strange things happen!! (Here there are a lot of squarings involved so this is what i felt!
11But shouldn't we be getting two values which r equal but opposite in sign?
11I found out where i went wrong. Thnx :)
P.S: you were spot on abt me finding 3 as a root ;p
It's all in the game... :))
62good that you asked...
I solved it by observation... like a puzzle.!!
And it was like a 30 second solution .. so i could be wrong!!
what i did was this
x=√7+√7-√7........
let
y = x=√7-√7........
so x=√7+y
and y=√7-x
the first thing that struck me was y=2 will give x=3 and x=3 gives y=2
I did not question anything.. Neither did i try to solve!!! It just struck in a flash!!
Where did u get this 3 as a root?? only bcos you had this solution in front of you that i gave.. (sorry for underestimating your observation [6]) But i dont think you could have found 3 as a root without knowing!!
I think in your question, x=-2 is also one root!! which i am getting bcos in taking squares, the signs get cancelled... so check from my logic if x=-2 is indeed a root!!
Dont tell me u were able to observe 3 as a root.. but not -2 ;P
11Let, x=√7+√7-√7........
= √7+√(7-x)
x2-7= √7-x
x4+49-14x2= 7-x
thus, x4-14x2+x+42=0
(x3+3x2-5x-14)(x-3)=0
pls check if this is right. If so, then then why is 3 the only solution?