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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