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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Updated on: 16 Nov 2016, 22:51
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Difficulty:

45% (medium)

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^2-nx+1), n=?

A. ±1/√2
B. ±1
C. ±√2
D. ±√3
E. ±2

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 29 Sep 2016, 05:23. Last edited by MathRevolution on 16 Nov 2016, 22:51, edited 2 times in total. Current Student Joined: 24 Jul 2016 Posts: 80 Location: United States (MI) GMAT 1: 730 Q51 V40 GPA: 3.6 Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink] Show Tags 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6826 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? [#permalink] Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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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.

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

Thanks
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Joined: 29 Jun 2016
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15 Nov 2016, 08:35
given ^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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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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15 Nov 2016, 08:58
3
1
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.

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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15 Nov 2016, 14:37
souvonik2k: Thank you!
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16 Nov 2016, 22:52
Hi all,

We have revised the question.
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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)(1^3-n+1)
Or (2+n)(2-n) = 2
Or 4-n^2 = 2
Or n = ±√2
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Re: If (x^4)+1=(x^2+nx+1)(x^2-nx+1), n=? &nbs [#permalink] 27 Feb 2017, 04:08
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