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Math Revolution GMAT Instructor
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If (x^4)+1=(x^2+nx+1)(x^2nx+1), n=?
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Updated on: 16 Nov 2016, 22:51
Question Stats:
69% (02:04) correct 31% (02:53) wrong based on 131 sessions
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If (x^4)+1=(x^2+nx+1)(x^2nx+1), n=? A. ±1/√2 B. ±1 C. ±√2 D. ±√3 E. ±2
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Re: If (x^4)+1=(x^2+nx+1)(x^2nx+1), n=?
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30 Sep 2016, 02:53
Answer is C. N can be +2 or 2. What's the OA? Sent from my iPhone using GMAT Club Forum mobile app



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Re: If (x^4)+1=(x^2+nx+1)(x^2nx+1), n=?
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01 Oct 2016, 01:06
==>If you substitute x=1, from 1^6+1=(1+n+1)(1n+1), you get 2=(2+n)(2n)=4n^2. Naturally, you get n^2=42=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^2nx+1), n=?
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15 Nov 2016, 07:44
MathRevolution wrote: ==>If you substitute x=1, from 1^6+1=(1+n+1)(1n+1), you get 2=(2+n)(2n)=4n^2. Naturally, you get n^2=42=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^2nx+1), n=?
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15 Nov 2016, 08:35
given ^6 + 1= (x^3+nx+1)(x^3nx+1) solve RHS: [(x^3+1)+nx][(x^3+1)nx] by using (a+b)(ab)=a^2b^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^2nx+1), n=?
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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^2nx+1), n=?
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15 Nov 2016, 08:58
WilDThiNg wrote: MathRevolution wrote: ==>If you substitute x=1, from 1^6+1=(1+n+1)(1n+1), you get 2=(2+n)(2n)=4n^2. Naturally, you get n^2=42=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 \((x^3+nx+1)(x^3nx+1) = (x^3+1)^2  (nx)^2\) using the formula \((a+b)(ab)= a^2b^2\) equal to , \(x^6+1+2x^3n^2x^2\) equating LHS=RHS \(x^6+1=x^6+1+2x^3n^2x^2\) gives \(2x^3n^2x^2=0\) therefore \(n^2=2x\) n=\(±\sqrt{2x}\) When x=0, n=0 when x=1, n=\(±\sqrt{2}\) which is option C If u found my post useful, press kudos!
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Re: If (x^4)+1=(x^2+nx+1)(x^2nx+1), n=?
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15 Nov 2016, 14:37



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