GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =
[#permalink]
18 Oct 2017, 22:48
18 Oct 2017, 22:48
00:00
Question Stats:
100% (01:08) correct
0% (00:00) wrong based on 11 sessions
Hide Show timer Statistics
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.
(A) -6
(B) -4
(C) -1/3
(D) 1/3
(E) 3
Kudos for correct solution.
Re: If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =
[#permalink]
19 Oct 2017, 00:28
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
Re: If (6 + 2/x)(x - 4) = 0, and x does not equal 4, then x =
[#permalink]
19 Oct 2017, 01:18
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.
(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 = \(-\frac{1}{3}\)
