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If 5 - 6/x = x, then x has how many possible values?

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If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 29 Dec 2012, 06:07
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If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number
[Reveal] Spoiler: OA
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 29 Dec 2012, 06:09
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 01 Jan 2013, 10:57
Bunuel wrote:
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

5 - 6/x = x --> (5x - 6)/x = x --> 5x -6 = x^2 --> (x-3)(x-2)=0 --> x=3 or x=2.

Answer: C.

That's cool Bunuel.
I can approach this kind of equation with Delta method. However I'm very interested in your factoring method.
Would you please share how you quickly factor the equation x^2 - 5x + 6 = 0 into (x-3)(x-2) = 0. Thanks
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 02 Jan 2013, 03:51
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akhandamandala wrote:
Bunuel wrote:
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

5 - 6/x = x --> (5x - 6)/x = x --> 5x -6 = x^2 --> (x-3)(x-2)=0 --> x=3 or x=2.

Answer: C.

That's cool Bunuel.
I can approach this kind of equation with Delta method. However I'm very interested in your factoring method.
Would you please share how you quickly factor the equation x^2 - 5x + 6 = 0 into (x-3)(x-2) = 0. Thanks


Solving and Factoring Quadratics:
http://www.purplemath.com/modules/solvquad.htm
http://www.purplemath.com/modules/factquad.htm

Hope it helps.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 02 Jan 2013, 15:30
that's great, it helps a lot. I may save 30 seconds from this method
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 23 Sep 2013, 13:27
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Since the question ask possible values, Can we apply following rules instead of factoring quadratic equation.
Calculate b^2 and 4ac

If b^2 > 4ac then 2 solutions
If b^2 = 4ac then 1 solution
If b^2 < 4ac then undefined

I think this method will also handle cases if GMAT choose to provide equations which can't be easily factored.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 02 Oct 2014, 02:30
Walkabout wrote:
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number


\(5 - \frac{6}{x} = x\)

\(x^2 - 5x + 6 = 0\)

This is a proper quadratic equation in the format \(ax^2 + bx + c = 0\)

This will provide 2 answers. No calculation required :)

Answer = C
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If 5 - 6/x = x, then x has how many possible values? [#permalink] New post 05 Apr 2015, 08:47
PareshGmat wrote:
Walkabout wrote:
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number


\(5 - \frac{6}{x} = x\)

\(x^2 - 5x + 6 = 0\)

This is a proper quadratic equation in the format \(ax^2 + bx + c = 0\)

This will provide 2 answers. No calculation required :)

Answer = C


You could still not have exactly two answers, depending on what the discriminant gives you, right? As I understand it: get the quadratic equation, calculate the discriminant, then determine how many possible answers you have. (In this case the answer is still C, two answers, but it's not due to the quoted reasoning).

Thoughts?
If 5 - 6/x = x, then x has how many possible values?   [#permalink] 05 Apr 2015, 08:47
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