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.
If 6-9/x = x, then x has how many...
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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.
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.
Regards,
Sathish
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Multiplying both sides by x, we get: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
6x - 9 = x²
0 = x² - 6x + 9
0 = (x-3)(x-3).
Thus, x=3.
The correct answer is B.
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Hi LUANDATO,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
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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Multiplying the given equation by x, we have:BTGmoderatorLU 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
The OA is B.
Can anye expert explain this PS question please? Thanks.
6x - 9 = x^2
x^2 - 6x + 9 = 0
(x - 3)(x - 3) = 0
x = 3
Answer: B
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Given: 6 - 9/x = xBTGmoderatorLU 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
Multiply both sides of the equation by x to get: 6x - 9 = x²
Rearrange to get: x² - 6x + 9 = 0
Factor to get: (x - 3) (x - 3) = 0
So, x = 3 (one solution)
Answer: B
Cheers,
Brent