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If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =

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If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x = [#permalink] New post 18 Oct 2017, 22:48
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If \((6 + \frac{2}{x})(x - 4) = 0\), and x does not equal 4, then x =

(A) -6
(B) -4
(C) -1/3
(D) 1/3
(E) 3




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[Reveal] Spoiler: OA
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Re: If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x = [#permalink] New post 19 Oct 2017, 00:28
Multiplying the two expressions in parenthesis we get \(6x-24-2-\frac{8}{x}=0\). This can be rewritten as \(6x^2-22x-8=0\). Solving it we get two solutions \(x = 4\) and \(x = -\frac{1}{3}\). Given that x cannot be equal to 4 the solution is the other one, answer C
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Re: If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x = [#permalink] New post 19 Oct 2017, 01:18
Bunuel wrote:
If \((6 + \frac{2}{x})(x - 4) = 0\), and x does not equal 4, then x =

(A) -6
(B) -4
(C) -1/3
(D) 1/3
(E) 3




Kudos for correct solution.


\((6 + \frac{2}{x})(x - 4) = 0\)

or we can write as = 6x^2 - 22x - 8 =0

or 3x^2 -11x - 4 = 0 (dividing by 2) which is in the form ax^2 + bx + c

Now we need the value of x , the best way in the complex equation is to use the formula, which I prefer

x= \((-b +- \sqrt{(b^2-4*a*c)}) /2*a\)

x= \((-11 +- \sqrt{169}) / 6\) (where a= 3, b= -11 c = -4)

or x = \(\frac{(11+- 13)}{6}\)

Now taking positive value
x= \(\frac{(11+13)}{6}\) = 4

Taking negative value

x= \(\frac{(11-13)}{6}\) = \(-\frac{1}{3}\)

Since x= 4 is not possible, therefore x = \(-\frac{1}{3}\)
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Re: If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =   [#permalink] 19 Oct 2017, 01:18
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If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =

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