TargetIIT

It is given that a polynomial with rational coefficients has n real roots.

One of them is 1+[sqrt]5[/sqrt]

Prove that 1-[sqrt]5[/sqrt] is another root!


Please realise that the implication of this are very huge and very very important in guessing second roots in IIT JEE examination....

Please try to prove and understand the above.

Hint: How do you prove that complex conjugates appear in pairs?


btw : do read post 17.. that was the question i was saying that got butchered!


[b]Mistake:[/b] Does x[p]3[/p]+2 have 2 irrational pair of roots? Well this holds only for even powers.

[b]Is the above mistake the end of it all?[/b] Well I got excited in posting this question. I was sure I wud manage a proof ;)
but the fact is that even if I dont.. then This is a must know for exams :)

I got a question today which i thought was very arbitrary.

This helped me in killing that one...

And I realised that this was very essential for a JEE aspirant!

have a doubt bhaiyya,
A  5th degree equation is given and if it is asked to find the number of real roots and complex roots,
how do we do it??

EVEN DIS :
[hide]I guess the fact z and z' having equal implications can b used

LET DA pOLY EQUAL to :

f(x) = (x-a)(x-b)(x-c)(x-d)......(x-n)


now if f(1+√5) = 0 = f(z)

Applying complement(alias conjugate [3]) on both sides,

We wud hav 0 = f(z') = f(1-√5)[/hide]

complement??

TOTALLY WRONG :

[hide]Wrong language

I meant : Z' is a conjugate(complement [3] ) of Z

And ya taking the polynomial as

f(x) = ax^n + bx^(n-1) + .............. (n-1)x + n
wud make more sense

coz then we can see dat

f(z) =  az^n + bz^(n-1) + .............. (n-1)z + n = 0

NOW taking conjugate on bot sides

f(z') = az'^n + ................ n = 0' = 0[/hide]

how do u use the spirit of conjugating in proving the question here?

let f(x)=a[ss]0[/ss]+a[ss]1[/ss]x+a[ss]2[/ss]x[p]2[/p] + ........+ a[ss]n[/ss]x[p]n[/p]

where all coeffiects are rational.

let the roots be b[ss]1[/ss], b[ss]2[/ss] .........,b[ss]n[/ss]

now, let b1 is irrational and all others are rational.

then,

b1+b2+....+bn = a[ss]n-1[/ss] which is rational.

contradiction!

so there must be another root which is conjugate to it... thus cancelling the iirational part.

now, its possible that b1=1+√5 and b[r]=2-√5
then a[n-1] becomes rational .

but b1.b2.b3........b[ss]n[/ss] = a[ss]0[/ss]

in this, LHS is irrational... and RHS is rational...

so again contradiction..

hence if a root is of the form b+√c , then another one shud be b-√c .

Ye bhi GALAT HAI

[hide]Taking this as given : f(z) =  az^n + bz^(n-1) + .............. (n-1)z + n = 0

wat I did was Take conjugate on both the sides which hav equal values.

so wat I got was
LHS :  z'^n + bz'^(n-1) + .............. (n-1)z' + n
RHS : 0' = 0

so Z' is also a root.........
so if Z = 1+√5
Z' = 1 - √5
[/hide]

OH MY GOD!!!!!

I hav bn making blunders [11]
I took 1 + √5i

and 1 - √5i all the TIME

sky,

good work i wud say.. really good work

but you missed out on one very small thing...

how do you eliminate possibilities like

1+[sqrt]5[/sqrt], -[sqrt]5[/sqrt]/2 and -[sqrt]5[/sqrt]/2



PS: I like your way of thinking.. very good.

sir post that ques for us plz.......

[b][i]how do you eliminate possibilities like

1+√5, -√5/2 and -√5/2[/i][/b]

wHILE TAKING product of roots this possibilty will be eliminated naa Sir.....

[i]"but b1.b2.b3........bn = a0
in this, LHS is irrational... and RHS is rational...
so again contradiction.."
[/i]

deepanshul, the question is right at the top ...

Sir, I guess he refers to :
[i]
"got a question today which i thought was very arbitrary.

This helped me in killing that one...

And I realised that this was very essential for a JEE aspirant!"[/i]

btw is da case neglection option given by me true?

yes i meant that ques ......  wich u murdered

bhaiya..
[b]
how do you eliminate possibilities like

1+√5, -√5/2 and -√5/2[/b]

if we multiply... it will give 5/4[1+√5] which is irrational..

a is rational ? given ?

@sky .. i only gave an example to show why your method is not fool proof!

Try to see how to make it more fool proof!

achha....

but Sir,

the eleimination by using the fact dat product of roots is a rational
and product of 2 nos. is rational only wen either they r purely rational OR of the form :

a-√b , a + √b

makes it full proof naa.....

how can you prove that no other product of 3 terms makes it irrational!

say

a-[sqrt]b[/sqrt] , [sqrt]c[/sqrt]-[sqrt]d[/sqrt] and [sqrt]e[/sqrt]-[sqrt]f[/sqrt]

how do you know that their product wont be rational?

and ha that purely rational case will b considered in SUM of roots :

√b, √c be the only two rational  roots then

√b + √c  hass 2 b zero

so ho gaya na full proof

let f(x) = (x-r1)(x-r2)(x-r3)...(x-rn)

now since d polynomial has rational co-efficients...

if r1 is irrational, there must be an another root (its conjugate) to counter it so dat d entire product gives rational co-efficients... hence for every irrational root, there must exist another root which is conjugate to it so dat d entire product is rational...

[hide]Proof of sky is almost convincing but since bhaiya has pointed out error in it, i thought

of sharing dis... else, i fully agree wid her...[/hide]

let me know if i'm wrong...!!!

oh okie, got ur point sir!

[12]

BTW who is :-)

Priyam kya?? [7]

and ha do take this into consideration Mr./Mrs  :-)

√b, √c be the only two rational  roots then
√b + √c  hass 2 b zero........... Sum of roots is zero

yeah... dis question can b disproved easily than prooving it...!!!

waiting for bhaiya's final answer...!!!

P.S.: tapan, i think u can read my name in my signature... [3]

Arrey I thought this was integration :P

oh vo dekha nahi tha khadeer bhai

Check the mistake in the question..

Re read it :) ;)

subhaanallah... now d question got its flavour... [1]

awesome to read n ask... difficult to solve...!!! [2] [3]

which question [7]

#17 question..

we can write the expression as:

[sqrt]a+[sqrt]a+x[/sqrt][/sqrt] = x

=> [sqrt]a+[sqrt]a+[sqrt]a+......[sqrt]a+x[/sqrt][/sqrt][/sqrt][/sqrt] = x

=> a + x = x[p]2[/p]

=> x[p]2[/p]-x-a =0

=> x = (-1 ± [sqrt]1+4a[/sqrt])/2

one root is given... the other one is the other one......

[b]Note this holds only for the below mentioned conditions...... not other fractional powers.... because for other powers...[/b]

as p(x)=(x-a+(b)[p]1/n[/p])(x-a-(b)[p]1/n[/p])=((x-a)[p]2[/p]-(b)[p]2/n[/p]) [b]so 2/n must be integer....[/b]...........................([b]a[/b])
or [b]n must be 2/I... I is some integer[/b]...
so for n like 2,2/2,2/3,2/4....,2/(-1),2/(-2).... above rule holds....


so for other n(i.e.) n≠2/I (I is integer..)

for other n, problem arises in [b]step(i)[/b].. as p(x) has irrational terms..........................from([b]a[/b])
so remainder λx+μ will have irrational terms...[/b](i.e.λ and μ will have irrational terms)hence in last [b]step (ii)[/b] we can't say λ=0 and μ=0..   (as we can't compare rational and irrational parts then...)..

so a+(b)[p]1/n[/p] being a root of f(x) having rational coefficients
[b]doesn't[/b] imply a-(b)[p]1/n[/p] to be root...[b]for n≠2/I[/b] ,I is integer

For question in post 17
[im]http://codecogs.izyba.com/gif.latex?x%3D%5Csqrt%7Ba+%5Csqrt%7Ba+x%7D%7D%20%2C%20a%5Cin%20%28-%5Cfrac%7B1%7D%7B4%7D%2C0%29[/im]

If one root is [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1+%5Csqrt%7B4a+1%7D%7D%7B2%7D[/im] then other root is...
_________________________________


[im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1-%5Csqrt%7B4a+1%7D%7D%7B2%7D[/im]
_________________________________________________

[hide] is also a root of equation... but when it is written in polynomial form... (as proof if for polynomial function..)

But in writing that we have to square... (twice) so case gain...

But we see in original equation x≥0

But for [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1-%5Csqrt%7B4a+1%7D%7D%7B2%7D[/im]≥0

[im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B4a+1%7D[/im]≥1

4a+1≥1
so 4a≥0 or a≥0 but a<0 so [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1-%5Csqrt%7B4a+1%7D%7D%7B2%7D[/im] is not a root of this equation..[/hide]

Now dividing the polynomial form of given equation...
[im]http://codecogs.izyba.com/gif.latex?x%5E%7B4%7D-2ax%5E%7B2%7D-x+a%5E%7B2%7D-a[/im]
by [im]http://codecogs.izyba.com/gif.latex?x%5E%7B2%7D-x-a[/im]

obtained since two roots are known of this polynomial equation

we get [im]http://codecogs.izyba.com/gif.latex?x%5E%7B2%7D+x+%281-a%29[/im]

which has [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B4a-3%7D[/im] as discriminant... and for it to be positive a≥3/4.. but 1/4<a<0
so only two real roots..

So Is the above proof correct [7]
agar upar ka sab proof sahi hai tab....

This doesn't depend on degree but n..
and n in 1/n is not degree of equation... because....

consider equation.. x[p]3[/p]+x[p]2[/p]-x=0
here n is 2  in.... a+(b)[p]1/n[/p] not 3(which is degree of equation so true for odd degrees also.. not only for even degree..)... so n=2 therefore... other root will be a-(b)[p]1/n[/p]..( n=2 here)

[b]so finally it is valid for square roots only...[/b]

utna n=2/I.. bla.. bla.. kiye... and finally only square root...
[im]http://www.artofproblemsolving.com/Forum/images/smiles/rotfl.gif[/im]..(khoda pahad nikla chooha...)
as a+(b)[p]I/2[/p]=a+(b[p]I[/p])[p]1/2[/p].... :D :D

@priyam..... u wrote:

[b]But we see in original equation x≥0

But for (1-[sqrt]1+4a[/sqrt])/2 ≥0

=> [sqrt]1+4a[/sqrt] ≥ 1[/b]

how[7] [7] [7]

(1-[sqrt]1+4a[/sqrt])/2 ≥0 => (1-[sqrt]1+4a[/sqrt]) >= 0

=> [sqrt]1+4a[/sqrt] <= 1

=> 1+4a <= 1

=> 4a <=0 !!

tum aur kuchh sochke woh likhe the kya [7]

   
skygirl

#36 Posted 10:04pm 18-03-09  
Re: 18-03-09 The other root?

BUT I FEEL ^^^^ is quite perfect!!!

Priyam then y the contradiction??

PL. read POST #17 completely b4 seeing this one!!

The queestion says -.25<a<0

THEN how  can v take it as a given that a will not irrational [7]

Coz if the one or more coeffecients of a polynomial are irrational then the very spirit of our proof is MURDERED........... AS WE WILL HAV solv it, widout predicting it as (1-[sqrt]4a + 1[/sqrt])/2

But ya the method of infinity which sky has applied seems quite geniune

@priyam.......

u have given a wonderful proof!

but i have one thing to say...

in your n=2/I .. let I=2/3 [b][LET][/b]

then n=3

and  a+b^1/n becomes a+b^1/3

[b]ofcourse this doesnt mean that a-b^1/3 will also be a root

only becoz..  a-b^1/3 is NOT the conjugate of a+b^1/3[/b]

conjugate is something like if z is a complex no and we multiply z with its conjugate i.e z' we get a real no.
similarly if p is a irrational no and we multily it wid p' we get a rational no.

but (a-b^1/3)(a+b^1/3) is never rational..

so (a-b^1/3) is not a conjugate of (a+b^1/3)

[b]rather if (a+b^1/3) is a root then some p'=R/(a+b^1/3) [R is some rational no.] has to be the other root.[/b]

(2)  for this to be true p<0and equal to the magnitude of p =[sqrt] q[/sqrt]
ya fir donno 0

(1)

opening binomially (a+[sqrt]b[/sqrt])[p]n[/p]=[p]n[/p]C[ss]0[/ss]([sqrt]b[/sqrt])[p]n[/p]   +    [p]n[/p]C[ss]1[/ss]a([sqrt]b[/sqrt])[p]n-1[/p]   +  [p]n[/p]C[ss]2[/ss]a[p]2[/p]([sqrt]b[/sqrt])[p]n-2[/p]........ .....         [p]n[/p]C[ss]n[/ss]a[p]n[/p]=  p+[sqrt]q[/sqrt]


(a-[sqrt]b[/sqrt])[p]n[/p]=[p]n[/p]C[ss]0[/ss]([sqrt]b[/sqrt])[p]n[/p]   -    [p]n[/p]C[ss]1[/ss]a([sqrt]b[/sqrt])[p]n-1[/p]   +  [p]n[/p]C[ss]2[/ss]a[p]2[/p]([sqrt]b[/sqrt])[p]n-2[/p]........ .....         [p]n[/p]C[ss]n[/ss]a[p]n[/p]=    [12][12][12]

any calc mistake [12]

@sky.. post 43..

[i]in your n=2/I .. let I=2/3 [LET] ....[/i] [b]I is integer.. so how 2/3 [/b][11]

[i]then n=3

and  a+b^1/n becomes a+b^1/3[/i]

post 36 wala proof sahi nahi hai kya.. :(