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

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01 Jan 2013, 11:57

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

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

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02 Jan 2013, 04:51

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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.

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

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

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05 Apr 2015, 09:47

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).

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

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24 Apr 2016, 10:46

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.

Bunnel - Can't there be non-integer values that can satisfy this equation ? I don't think the question mentions that x has to be an integer. Please advise.

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

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24 Apr 2016, 11:24

Expert's post

Keysersoze10 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.

Bunnel - Can't there be non-integer values that can satisfy this equation ? I don't think the question mentions that x has to be an integer. Please advise.

We solved 5 - 6/x = x and got two solutions: 3 and 2. So, the answer to your question is no, this equation does not have any other solution, integer or non-integer. _________________

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

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24 Apr 2016, 11:39

Bunuel wrote:

Keysersoze10 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.

Bunnel - Can't there be non-integer values that can satisfy this equation ? I don't think the question mentions that x has to be an integer. Please advise.

We solved 5 - 6/x = x and got two solutions: 3 and 2. So, the answer to your question is no, this equation does not have any other solution, integer or non-integer.

Thanks Bunnel. I might be overthinking, but if we draw a parabola for the above equation wont there be infinite number of points on that parabola which would satisfy the above equation ?

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

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24 Apr 2016, 11:50

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Keysersoze10 wrote:

Thanks Bunnel. I might be overthinking, but if we draw a parabola for the above equation wont there be infinite number of points on that parabola which would satisfy the above equation ?

No. There will be only TWO. There is a clear solution given above, which gives TWO values of x satisfying the equation. _________________

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

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24 Apr 2016, 11:56

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Keysersoze10 wrote:

We solved 5 - 6/x = x and got two solutions: 3 and 2. So, the answer to your question is no, this equation does not have any other solution, integer or non-integer.

Thanks Bunnel. I might be overthinking, but if we draw a parabola for the above equation wont there be infinite number of points on that parabola which would satisfy the above equation ?

GMAT Quants is simple, keep it simple !!

For scoring 700+ and a good score in QA knowledge of Quadratic equation is more than sufficient, don't go into such depths , GMAT will seldom go into such depths....

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

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24 Apr 2016, 12:26

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Expert's post

Keysersoze10 wrote:

Thanks Bunnel. I might be overthinking, but if we draw a parabola for the above equation wont there be infinite number of points on that parabola which would satisfy the above equation ?

If you rearrange the given equation, 5-6/x = x --> (5-x) = 6/x means that you have to find points of intersection for the line y=5-x and the hyperbola y=6/x (do note that hyperbolas are asymptotic to the coordinate axes). Clearly, the line y=5-x can only intersect the hyperbola at 2 and only 2 points (refer to the image below):

Attachment:

2016-04-24_15-45-43.jpg [ 43.11 KiB | Viewed 1110 times ]

Thus you get 2 solutions.

FYI, knowledge of hyperbolas is not in GMAT's scope. _________________

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

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28 Apr 2016, 19:31

5 - (6/x) = x 5=x + (6/x) 5 = (x^2+6)/x 5x = x^2+6 <<<<< you can really stop here if you recognize this is a factorization problem and not a perfect square 0 = x^2 - 5x + 6 0 = (x-3)(x-2) 2 answers

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