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

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New post 09 Jan 2018, 02:54
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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. \(2\)
[Reveal] Spoiler: OA

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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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New post 09 Jan 2018, 06:11
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
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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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New post 09 Jan 2018, 10:21
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
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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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New post 11 Jan 2018, 00:44
=>

\(n^4 – 2^{4k} = 0\)
\(⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0\)
\(⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0\)
\(⇔ (n+2^k)(n-2^k) = 0\)
\(⇔ n = 2^k\)
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
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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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New post 25 Jan 2018, 22:41
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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New post 25 Jan 2018, 23:34
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
Re: Suppose x=-1 and k is a positive number. If n is a positive root of t   [#permalink] 25 Jan 2018, 23:34
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