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If 5 - 6/x = x, then x has how many possible values?
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Walkabout
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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 \(5 - \frac{6}{x} = x\), then x has how many possible values?

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

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

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

\(5x -6 = x^2\);

\((x-3)(x-2)=0\);

\(x=3\) or \(x=2\).

Answer: C.
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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  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
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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  02 Oct 2014, 02:30
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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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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  05 Apr 2015, 08:47
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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?
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  24 Apr 2016, 09:46
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.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  24 Apr 2016, 10:24
Expert Reply
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.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  24 Apr 2016, 10:39
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 ?
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  24 Apr 2016, 10:50
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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.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  24 Apr 2016, 10:56
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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....

Regards

AbhisheK
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  Updated on: 24 Apr 2016, 11:48
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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 ?



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
2016-04-24_15-45-43.jpg [ 43.11 KiB | Viewed 27031 times ]


Thus you get 2 solutions.

FYI, knowledge of hyperbolas is not in GMAT's scope.
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Originally posted by ENGRTOMBA2018 on 24 Apr 2016, 11:26.
Last edited by ENGRTOMBA2018 on 24 Apr 2016, 11:48, edited 2 times in total.
Edited the solution.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  28 Apr 2016, 18:31
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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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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  15 Jul 2016, 03:51
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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


We need to simplify the equation 5 – 6/x = x. We start by multiplying the entire equation by x and we obtain:

5x – 6 = x^2

x^2 – 5x + 6 = 0

(x – 3)(x – 2) = 0

x = 3 or x = 2.

We see that x has 2 possible values.

The answer is C.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  14 May 2017, 02:09
As its a quadric equation, there can be 2 values of x. So do you really to calculate or can select answer 2 without calculating what is the value of x?
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  14 May 2017, 02:18
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sisirkant wrote:
As its a quadric equation, there can be 2 values of x. So do you really to calculate or can select answer 2 without calculating what is the value of x?


Quadratic equation can have 0, 1 or two roots, so you should do some additional math here.
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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  17 Mar 2018, 11:03
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Hi All,

You'll have to do a bit of algebra to deduce the number of solutions to this equation…

5 - (6/x) = x

First, multiply everything by x…

5x - 6 = x^2

Now, move everything "to the right"….

0 = x^2 - 5x + 6

You can now factor this into two terms…

0 = (x -2)(x - 3)

And answer the question… There are 2 solutions

Final Answer:
Show: ::
C


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Re: If 5 - 6/x = x, then x has how many possible values? [#permalink] New post  28 Mar 2018, 09:43
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

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

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

\(5x -6 = x^2\);

\((x-3)(x-2)=0\);

\(x=3\) or \(x=2\).

Answer: C.


Well i wanted to reconfirm , we eliminated the possibility of x=0 , because equating x=0 in the equation doesn't solve it , Right ??
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