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# If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 =

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Math Expert
Joined: 02 Sep 2009
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 03:47
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Question Stats:

73% (01:27) correct 27% (01:49) wrong based on 60 sessions

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If x ≠ 0 and $$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$, then x^2 – x + 4 =

A. 5
B. 4
C. 3
D. 2
E. 1
[Reveal] Spoiler: OA

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 04:02
Solving the equation 1/x=x/x+1, further solving, the equation becomes x^2=x+1. So x^2-x=1, x(x-1)=1. So either x=1 or x-1=1 so either x=1 or x=2. Substituting the values in the equation, x^2-x+4= either 4 or 6. Since 6 is not an option so i am assuming answer is B i.e. 4. Is the logic correct bunuel

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 04:22
Solving the equation 1/x = 1/(1+1/x), we get x^2 = x+1. Substitute x^2 in equation x^2 - x+4. => x+1 - x+4 = 5. So option A.

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 04:45
Oh my bad...i missed a major detail. Correct is A

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 05:05
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Bunuel wrote:
If x ≠ 0 and $$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$, then x^2 – x + 4 =

A. 5
B. 4
C. 3
D. 2
E. 1

Hi...
$$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}...........\frac{1}{x}=\frac{x}{x+1}.......\frac{1}{x}-\frac{x}{x+1}=0........\frac{x+1-x*x}{x(x+1)}=0$$
So $$x^2-x-1=0$$ but $$x^2-x+4=x^2-x-1+5=0+5=5$$
A
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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17 Sep 2017, 07:17
Bunuel wrote:
If x ≠ 0 and $$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$, then x^2 – x + 4 =

A. 5
B. 4
C. 3
D. 2
E. 1

1/x = x/x+1 i.e; x^2-x-1=0;

We need to find the value of x^2-x+4 i.e; x^2-x-1+5 =0+5 =5

Ans:A

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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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25 Sep 2017, 03:30
chetan2u wrote:
Bunuel wrote:
If x ≠ 0 and $$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$, then x^2 – x + 4 =

A. 5
B. 4
C. 3
D. 2
E. 1

Hi...
$$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}...........\frac{1}{x}=\frac{x}{x+1}.......\frac{1}{x}-\frac{x}{x+1}=0........\frac{x+1-x*x}{x(x+1)}=0$$
So $$x^2-x-1=0$$ but $$x^2-x+4=x^2-x-1+5=0+5=5$$
A

Would you be able to expand on this a little more? I'm sorry, I'm just having a tough time following your method.

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Math Expert
Joined: 02 Sep 2009
Posts: 41875

Kudos [?]: 128463 [0], given: 12173

Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = [#permalink]

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25 Sep 2017, 03:46
imranmah wrote:
chetan2u wrote:
Bunuel wrote:
If x ≠ 0 and $$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$, then x^2 – x + 4 =

A. 5
B. 4
C. 3
D. 2
E. 1

Hi...
$$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}...........\frac{1}{x}=\frac{x}{x+1}.......\frac{1}{x}-\frac{x}{x+1}=0........\frac{x+1-x*x}{x(x+1)}=0$$
So $$x^2-x-1=0$$ but $$x^2-x+4=x^2-x-1+5=0+5=5$$
A

Would you be able to expand on this a little more? I'm sorry, I'm just having a tough time following your method.

$$\frac{1}{x}=\frac{1}{1+\frac{1}{x}}$$

Cross-multiply: $$1+\frac{1}{x} = x$$;

Multiply by x: $$x + 1 = x^2$$;

Re-arrange: $$x^2 - x = 1$$.

Therefore, $$x^2 – x + 4 = 1 + 4 = 5$$.

Hope it's clear.
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Re: If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 =   [#permalink] 25 Sep 2017, 03:46
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