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

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Math Expert
Joined: 02 Sep 2009
Posts: 44423
If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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25 Sep 2017, 23:35
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60% (01:16) correct 40% (01:46) wrong based on 50 sessions

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

(1) x – x^y = –2
(2) y = –2
[Reveal] Spoiler: OA

_________________
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1018
Location: India
GPA: 3.82
If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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26 Sep 2017, 02:44
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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 $$x=-1$$
Director
Joined: 21 Mar 2016
Posts: 549
Re: If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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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????
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1018
Location: India
GPA: 3.82
Re: If x and y are negative integers such that x = x^y + y, what is x? [#permalink]

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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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