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# Suppose x=-1 and k is a positive number. If n is a positive root of t

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Math Revolution GMAT Instructor
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Suppose x=-1 and k is a positive number. If n is a positive root of t  [#permalink]

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09 Jan 2018, 03:54
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5% (low)

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86% (02:09) correct 14% (02:00) wrong based on 74 sessions

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[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Expert Joined: 02 Aug 2009 Posts: 7952 Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink] ### Show Tags 09 Jan 2018, 07:11 MathRevolution wrote: [GMAT math practice question] Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$? A. $$-2$$ B. $$-1$$ C. $$0$$ D. $$1$$ E. $$2$$ FIRST step - solve for 'n'... $$n^4 – 2^{4k} = 0..........(n^2-2^{2k})(n^2+2^{2k})=0...............(n-2^{k})(n+2^k)(n^2+2^{2k})=0$$ so what all can be the value of n... $$2^k$$ or $$-(2^k)$$ or $$\sqrt{-(2^{2k})}$$, which is not possible positive root of n is $$2^k$$ and since k is positive integer, $$2^k$$ or n will also be EVEN.. $$x+x^n+x^{n+1}+x^{n+2} = (-1)+(-1)^{even}+(-1)^{even+1}+(-1)^{even+2}=(-1)+1+(-1)+1=0$$ C _________________ Retired Moderator Joined: 25 Feb 2013 Posts: 1180 Location: India GPA: 3.82 Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink] ### Show Tags 09 Jan 2018, 11:21 1 1 MathRevolution wrote: [GMAT math practice question] Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$? A. $$-2$$ B. $$-1$$ C. $$0$$ D. $$1$$ E. $$2$$ $$n^4-2^{4k}=0=>n^4=2^{4k}$$. Taking fourth root of both sides we have $$n=2^k=Even$$ as $$k$$ is a positive integer(Note as $$n$$ is a positive root, so $$n=-2^k$$ is not possible here) Hence $$x+x^n+x^{n+1}+x^{n+2}=-1+(-1)^{even}+(-1)^{odd}+(-1)^{even}=0$$ Option C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8001 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink] ### Show Tags 11 Jan 2018, 01:44 => $$n^4 – 2^{4k} = 0$$ $$⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0$$ $$⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0$$ $$⇔ (n+2^k)(n-2^k) = 0$$ $$⇔ n = 2^k$$ Since$$k > 0, n$$ must be even. Thus $$n = 2m$$ for some integer $$m$$. It follows that $$x+x^n+x^{n+1}+x^{n+2}$$ $$= (-1)+(-1)^n+(-1)^{n+1}+(-1)^{n+2}$$ $$= (-1)+(-1)^{2m}+(-1)^{2m} +1+(-1)^{2m+2}$$ $$= (-1) + 1 + (-1) + 1 = 0$$ 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$79 for 1 month Online Course"
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t  [#permalink]

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25 Jan 2018, 23:41
MathRevolution wrote:
=>

$$n^4 – 2^{4k} = 0$$
$$⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0$$
$$⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0$$
$$⇔ (n+2^k)(n-2^k) = 0$$
$$⇔ n = 2^k$$
Since$$k > 0, n$$ must be even.

Thus $$n = 2m$$ for some integer $$m$$. It follows that
$$x+x^n+x^{n+1}+x^{n+2}$$
$$= (-1)+(-1)^n+(-1)^{n+1}+(-1)^{n+2}$$
$$= (-1)+(-1)^{2m}+(-1)^{2m} +1+(-1)^{2m+2}$$
$$= (-1) + 1 + (-1) + 1 = 0$$

Therefore, the answer is C.
Answer: C

I do not understand why do we need this complex calculation, when we know it for a fact that any No. multiplied by 2 is always an Even number and as "K" is positive, so 2^4k has to be an even no. and thus n^4 is even too. and that makes the process much easier.
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t  [#permalink]

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26 Jan 2018, 00:34
MathRevolution wrote:
[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$

Hi MathRevolution

shouldn't it be better to mentioned "k" as positive integer rather than positive number because positive number can be a decimal value which will create ambiguity.

we all have solved the question assuming "k" to be a positive integer
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t  [#permalink]

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24 May 2019, 12:46
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t   [#permalink] 24 May 2019, 12:46
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# Suppose x=-1 and k is a positive number. If n is a positive root of t

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