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555-605 (Medium)|   Algebra|      
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Bunuel
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 03:47
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

Question Stats:

73% (01:38) correct 27% (02:09) wrong based on 128 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
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A
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kaulshanx
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 04:02
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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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madhubala0792
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 04:22
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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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kaulshanx
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 04:45
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Oh my bad...i missed a major detail. Correct is A

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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 05:05
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Bunuel
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
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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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Arsh4MBA
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 17 Sep 2017, 07:17
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Bunuel
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
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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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imranmah
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 25 Sep 2017, 03:30
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chetan2u
Bunuel
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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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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Bunuel
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 25 Sep 2017, 03:46
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imranmah
chetan2u
Bunuel
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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\(\frac{1}{x}=\frac{1}{1+\frac{1}{x}}\)

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

Multiply by x: \(x + 1 = x^2\);

Re-arrange: \(x^2 - x = 1\).

Therefore, \(x^2 – x + 4 = 1 + 4 = 5\).

Answer: A.

Hope it's clear.
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If x ≠ 0 and 1/x = 1/(1 + 1/x}, then x^2 – x + 4 = 20 Dec 2023, 07:33
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Bunuel
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
Show more
\(\frac{1}{x}=\frac{1}{1+\frac{1}{x}}\)

Or, \(\frac{1}{x}=\frac{1}{\frac{ x+1}{x}}\)

Or, \(\frac{1}{x}=\frac{x}{(x + 1)}\)

Or, \(x^2 = x + 1\)

Or,\(x^2 - x = 1\)

\((x^2 – x ) + 4 = 1 + 4 => 5\) , Answer must be (A)
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