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Manager  Joined: 02 Dec 2012
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If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58442
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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3
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 $$x=2$$.

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Intern  Joined: 10 Sep 2013
Posts: 1
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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5
1
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.
General Discussion
Intern  Joined: 17 Nov 2012
Posts: 17
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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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Joined: 02 Sep 2009
Posts: 58442
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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2
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

Solving and Factoring Quadratics:

Hope it helps.
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Intern  Joined: 17 Nov 2012
Posts: 17
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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]

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3
1
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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Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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

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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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58442
Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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.

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]

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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.

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]

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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]

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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]

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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 [ 43.11 KiB | Viewed 19634 times ]

Thus you get 2 solutions.

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

Originally posted by ENGRTOMBA2018 on 24 Apr 2016, 12:26.
Last edited by ENGRTOMBA2018 on 24 Apr 2016, 12:48, 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]

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1
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)
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Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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

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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GMAT 1: 550 Q44 V22 Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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]

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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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If 5 - 6/x = x, then x has how many possible values?  [#permalink]

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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

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

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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 $$x=2$$.

Well i wanted to reconfirm , we eliminated the possibility of x=0 , because equating x=0 in the equation doesn't solve it , Right ?? Re: If 5 - 6/x = x, then x has how many possible values?   [#permalink] 28 Mar 2018, 10:43

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