Home » Ask & Discuss » Mathematics
« Back to Discussion

Algebra

Blazing goIITian

Joined:	22 Jan 2007
Post:	2039

27 Jan 2007 21:16:17 IST

0 People liked this

993

Quadratic

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

If , are the roots of x²+px +q=0 and also of x²ⁿ+pⁿqⁿ+qⁿ=0 and if / ,/ are the roots of xⁿ+1 +(x+1)ⁿ=0,then n is:

(a) an odd integer

(b) an even integer

(c) any integer

(d) none

Comments (11)

puneet

Forum Expert

Joined: 19 Oct 2006

Posts: 1966

28 Jan 2007 00:20:40 IST

Like 0 people liked this

hiiiii Neeraj ..

The question has not come up properly due to some problem .. u know these crosses.. what is in there place .. it is not clear .

I was wondering if u can post this qusetion again ..

I hope u will not mind ...

cheers

Neeraj Agarwal

Blazing goIITian

Joined: 22 Jan 2007

Posts: 2039

16 Feb 2007 13:41:32 IST

Like 0 people liked this

alpha ,beta, alpha/beta ,beta/alpha

vash

Cool goIITian

Joined: 10 Feb 2007

Posts: 53

16 Feb 2007 13:44:02 IST

Like 0 people liked this

would u plz put it once again i am not gettin u

vash

Cool goIITian

Joined: 10 Feb 2007

Posts: 53

16 Feb 2007 14:06:17 IST

Like 0 people liked this

ok i got ur problem

i think in the second equation it's xⁿinstead of pⁿ

= -p

= q

the next equation has roots

ⁿ &

ⁿ

therefore

ⁿ +

ⁿ= -p ( let xⁿ=y ) ok

ⁿ=k and

ⁿ= m ( disgustin typin all that)

km=qⁿ

now (

)^n + (

+1)^n + 1 (since

is its root)

=k/m + (

)^n/

^n + 1 ( where k and m hav their usual meanings)

takin l.c.m

= {-p^n + (-p)^n}/

ⁿ

now if it is a factor then the above term should add upto zero therefore n is even

now that was a tough job to pen down all those xponentials huh..... plzzzzzzz give me rates if i deserve.

if u didn't get this i am there to sort it out

Neeraj Agarwal

Blazing goIITian

Joined: 22 Jan 2007

Posts: 2039

17 Feb 2007 15:40:56 IST

Like 0 people liked this

But why are the roots of the second equation alpha^n and beta^n?

Its roots are given alpha and beta?I've understood the rest of the solution

Your answer is correct
thanks

Ramya Hegde

Blazing goIITian

Joined: 8 Feb 2007

Posts: 612

17 Feb 2007 15:51:11 IST

Like 0 people liked this

since

and

are the roots of the second equation, hence the roots of "y"(=xⁿ) must be equal to (alpha)ⁿ and (beta)ⁿ, because y=xⁿ, and x=alpha,beta.

vash

Cool goIITian

Joined: 10 Feb 2007

Posts: 53

17 Feb 2007 16:58:55 IST

Like 0 people liked this

ya i agree with ramya by symmetry if

are roots of equation x^2 + ax + c=0, then

^n &

^n should b the roots of the other equation.

Neeraj Agarwal

Blazing goIITian

Joined: 22 Jan 2007

Posts: 2039

17 Feb 2007 18:21:51 IST

Like 0 people liked this

WHY???

If first polynomial is represented by 'y' ,then the other equation is not
y^n .

then where comes the symmetry??

Ramya Hegde

Blazing goIITian

Joined: 8 Feb 2007

Posts: 612

17 Feb 2007 20:47:22 IST

Like 1 people liked this

but u r taking xⁿ=y and when u substitute this in the second equation, u get two roots of y,mind u, not of xⁿ, and u equate this to (alpha)ⁿ and (beta)ⁿ, bcoz y=xⁿ, and from the first equation, we know, x= alpha and beta

Neeraj Agarwal

Blazing goIITian

Joined: 22 Jan 2007

Posts: 2039

17 Feb 2007 21:16:47 IST

Like 0 people liked this

I got it....thanks

vash

Cool goIITian

Joined: 10 Feb 2007

Posts: 53

18 Feb 2007 13:57:04 IST

Like 0 people liked this

ok let me put it the other way

see alpha and beta are the roots of the second equation so alpha^2n + p^n*alpha^n +q^n =0 since it satisfies it. now this is a quadratic with roots alpha^n & beta^n

. so the roots are alpha^n and beta^n. i think u got my point!!!!!!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team