1[3] thanx !! aaj raat so payenge thoda [3]
kal 4 baje uthke post kii thi... aur poori raat wahi dkhi sapne me [3]
It is given that a polynomial with rational coefficients has n real roots.
One of them is 1+√5
Prove that 1-√5 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!
Mistake: Does x3+2 have 2 irrational pair of roots? Well this holds only for even powers.
Is the above mistake the end of it all? 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 :)
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61 Answers
1
62-\frac{1}{4}<a<0
\sqrt{a+\sqrt{a+x}}=x
one root is \frac{1+\sqrt{4a+1}}{2}
Find the other root...
To be true it has nothing to do with this question, but for an objective paper, I guess you guys know how brutally this wud get killed ;)
1have 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??
21EVEN DIS :
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)
21TOTALLY WRONG :
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
62how do u use the spirit of conjugating in proving the question here?
1let f(x)=a0+a1x+a2x2 + ........+ anxn
where all coeffiects are rational.
let the roots be b1, b2 .........,bn
now, let b1 is irrational and all others are rational.
then,
b1+b2+....+bn = an-1 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........bn = a0
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 .
21Ye bhi GALAT HAI
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
21OH MY GOD!!!!!
I hav bn making blunders [11]
I took 1 + √5i
and 1 - √5i all the TIME
62sky,
good work i wud say.. really good work
but you missed out on one very small thing...
how do you eliminate possibilities like
1+√5, -√5/2 and -√5/2
PS: I like your way of thinking.. very good.
21how do you eliminate possibilities like
1+√5, -√5/2 and -√5/2
wHILE TAKING product of roots this possibilty will be eliminated naa Sir.....
"but b1.b2.b3........bn = a0
in this, LHS is irrational... and RHS is rational...
so again contradiction.."
21Sir, I guess he refers to :
"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!"
btw is da case neglection option given by me true?
62I 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!
1bhaiya..
how do you eliminate possibilities like
1+√5, -√5/2 and -√5/2
if we multiply... it will give 5/4[1+√5] which is irrational..
62@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!
21but 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.....
62how can you prove that no other product of 3 terms makes it irrational!
say
a-√b , √c-√d and √e-√f
how do you know that their product wont be rational?
21and 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
1let 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...
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...
let me know if i'm wrong...!!!
21BTW 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
1yeah... 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]