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If x and y are negative integers such that x = x^y + y, what is x?

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V
Joined: 02 Sep 2009
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If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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New post 25 Sep 2017, 23:35
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D
E

Difficulty:

  65% (hard)

Question Stats:

62% (01:19) correct 38% (01:46) wrong based on 52 sessions

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Joined: 25 Feb 2013
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If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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New post 26 Sep 2017, 02:44
1
Bunuel wrote:
If x and y are negative integers such that x = x^y + y, what is x?

(1) x – x^y = –2
(2) y = –2


\(x = x^y + y\) or \(x-x^y = y\)

Statement 1: this implies that \(y = -2\). putting the value of \(y\) in question stem equation we get

\(x-x^{-2}=-2\)

\(x-\frac{1}{x^2} = -2\), as \(x\) is a negative integer, only \(x=-1\) satisfies this equation. Hence sufficient

Statement 2: Directly provides the value of \(y\). As explained above only \(x=-1\) satisfies the equation. Hence sufficient

Option D

---------------------------------------------------

Factorization to find the value of \(x\)

\(x-\frac{1}{x^2} = -2\) or \(x^3+2x^2-1=0\)

here \(f(x)=x^3+2x^2-1\), Notice that \(f(-1)=0\), or \(x=-1\) is a root of the equation, so \(x+1=0\) is a factor of the equation.

Divide the equation by \((x+1)\) to get \(x^3+2x^2-1=(x+1)(x^2+x-1) =0\),

Now \(x\) is a negative integer, so for any value of \(x\), \(x^2+x-1\) will not be \(0\)

Hence \(x+1=0\) or \(x=-1\)
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Re: If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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New post 26 Sep 2017, 10:56
niks18 wrote:
Bunuel wrote:
If x and y are negative integers such that x = x^y + y, what is x?

(1) x – x^y = –2
(2) y = –2


\(x = x^y + y\) or \(x-x^y = y\)

Statement 1: this implies that \(y = -2\). putting the value of \(y\) in question stem equation we get

\(x-x^{-2}=-2\)

\(x-\frac{1}{x^2} = -2\), as \(x\) is a negative integer, only \(x=-1\) satisfies this equation. Hence sufficient

Statement 2: Directly provides the value of \(y\). As explained above only \(x=-1\) satisfies the equation. Hence sufficient

Option D

---------------------------------------------------

Factorization to find the value of \(x\)

\(x-\frac{1}{x^2} = -2\) or \(x^3+2x^2-1=0\)

here \(f(x)=x^3+2x^2-1\), Notice that \(f(-1)=0\), or \(x=-1\) is a root of the equation, so \(x+1=0\) is a factor of the equation.

Divide the equation by \((x+1)\) to get \(x^3+2x^2-1=(x+1)(x^2+x-1) =0\),

Now \(x\) is a negative integer, so for any value of \(x\), \(x^2+x-1\) will not be \(0\)

Hence \(x+1=0\) or \(x=-1\)



any easier way to solve this????
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Re: If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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New post 26 Sep 2017, 11:01
Quote:
any easier way to solve this????


Hi mohshu

I find this to be a very simple process as both the statements directly provide the value of y.
and since x is negative only one value of x satisfies the given equation.

Is there any specific question that i can address? and if you are finding the factorization process confusing, then that is just an FYI
Re: If x and y are negative integers such that x = x^y + y, what is x?   [#permalink] 26 Sep 2017, 11:01
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