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If 6-9/x = x, then x has how many...

This topic has 2 expert replies and 1 member reply

If 6-9/x = x, then x has how many...

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If $$6-\frac{9}{x}=x$$
then x has how many possible values?

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

The OA is B.

Can anye expert explain this PS question please? Thanks.

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6-9/x = x

When x=1 --> 6-9 = -3
When x=2 --> 6-4.5 = 1.5
When x=3 --> 6-3 = 3
When x=4 --> 6-9/4 not equal to 4 and same is the case for any other value for x henceforth

So x has only one possible value 3.

Other way to solve: Algebrically, this will be of form (x-a)(x-b) = x$$^2$$ -ax -bx +ab
From the question, ab = 9. Since 3*3 = 9. a=b=3. Hence the solution is (x-3)$$^2$$ and x value will be 3 --> only one possible value.

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GMAT/MBA Expert

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LUANDATO wrote:
If $$6-\frac{9}{x}=x$$
then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) Three
Multiplying both sides by x, we get:
6x - 9 = x²
0 = x² - 6x + 9
0 = (x-3)(x-3).
Thus, x=3.

The correct answer is B.

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GMAT/MBA Expert

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Quote:
If $$6-\frac{9}{x}=x$$
then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) Three
Hi LUANDATO,
Let's take a look at your question.


$$6-\frac{9}{x}=x$$
$$6=x+\frac{9}{x}$$
Simplify right hand side by adding up like terms.
$$6=\frac{x^2+9}{x}$$
$$6x=x^2+9$$
$$x^2-6x+9=0$$
Let's factorize the left hand side by grouping,
$$x^2-3x-3x+9=0$$
$$x(x-3)-3(x-3)=0$$
$$(x-3)(x-3)=0$$
$$Either (x-3)=0 or (x-3)=0$$
It gives us:
$$x = 3$$

It means x has only one value.
Therefore, Option B is correct.

Hope this helps.
I am available if you'd like any follow up.

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