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If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=?

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If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=?

A. ±1/√2
B. ±1
C. ±√2
D. ±√3
E. ±2
[Reveal] Spoiler: OA

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Last edited by MathRevolution on 16 Nov 2016, 22:51, edited 2 times in total.
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 30 Sep 2016, 02:53
Answer is C. N can be +2 or -2. What's the OA?


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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 01 Oct 2016, 01:06
==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 07:44
MathRevolution wrote:
==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C


Hi, please can you show how to solve it alegebraically?

Thanks
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 08:35
given :x^6 + 1= (x^3+nx+1)(x^3-nx+1)
solve RHS:
[(x^3+1)+nx][(x^3+1)-nx]

by using (a+b)(a-b)=a^2-b^2

we have: a= (x^3+1) and b = nx

after solving we get: (x^3+1)^2 - (nx)^2
x^6+1+2x^3-(nx)^2
but on LHS we have x^6+1 so on RHS remaining terms should be zero. i.e. 2x^3 - (nx)^2 = 0.

so what should be next ? can any one tell me ?
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 08:48
guys please see my solution and put the given answer (underroot 2) in final equation:
x^6 +1 + 2x^3 - 2x^2.
so for RHS to be equal to LHS there should be 2x^2 instead of 2x^3. so i think that there is some mistake in question.
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 08:58
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WilDThiNg wrote:
MathRevolution wrote:
==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C


Hi, please can you show how to solve it alegebraically?

Thanks


Solving the Right hand side which is \((x^3+nx+1)(x^3-nx+1) = (x^3+1)^2 - (nx)^2\) ---using the formula \((a+b)(a-b)= a^2-b^2\)
equal to , \(x^6+1+2x^3-n^2x^2\)
equating LHS=RHS
\(x^6+1=x^6+1+2x^3-n^2x^2\)
gives \(2x^3-n^2x^2=0\)
therefore \(n^2=2x\)
n=\(±\sqrt{2x}\)
When x=0, n=0
when x=1, n=\(±\sqrt{2}\) which is option C

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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 09:28
ah... thanx. :)
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 15 Nov 2016, 14:37
souvonik2k: Thank you!
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 16 Nov 2016, 22:52
Hi all,

We have revised the question.
Sorry about the confusion.
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink]

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New post 27 Feb 2017, 04:08
Substitute x = 1 in the equation. We will have
1^6+1=(1^3+n+1)(1^3-n+1)
Or (2+n)(2-n) = 2
Or 4-n^2 = 2
Or n = ±√2
Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=?   [#permalink] 27 Feb 2017, 04:08
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