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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]  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
Math Expert
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink]  29 Dec 2012, 06:09
Expert's post
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.

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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink]  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.

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]  02 Jan 2013, 03:51
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Expert's post
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.

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

Hope it helps.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink]  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]  23 Sep 2013, 13:27
2
KUDOS
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]  02 Oct 2014, 02:30
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

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

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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