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04 Mar 2020, 02:31
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:56) correct
40%
(02:39)
wrong
based on 171
sessions
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If \(6(x + \frac{1}{x})^2 − 35(x + \frac{1}{x}) + 50 = 0\), what could be the values of x
A. 3/10 or 2/5
B. 3/10 or 5/2
C. 10/3 or 2/5
D. 1/3 or 5/2
E. 1/2 or 3
Are You Up For the Challenge: 700 Level Questions
A. 3/10 or 2/5
B. 3/10 or 5/2
C. 10/3 or 2/5
D. 1/3 or 5/2
E. 1/2 or 3
Are You Up For the Challenge: 700 Level Questions
04 Mar 2020, 03:01
Kudos
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assume \(x+\frac{1}{x} = k\)
\(6k^2-35k+50 = 0\)
\(6k^2- 15k -20k + 50=0\)
\(3k(2k-5) -10(2k-5) =0\)
k= \(\frac{10}{3}\) or \(\frac{5}{2}\)
If \(x+\frac{1}{x}= \frac{10}{3} = 3+\frac{1}{3}\)
x= 3 or \(\frac{1}{3}\)
\(x+\frac{1}{x }= \frac{5}{2} = 2+\frac{1}{2}\)
x = 2 or \(\frac{1}{2}\)
\(6k^2-35k+50 = 0\)
\(6k^2- 15k -20k + 50=0\)
\(3k(2k-5) -10(2k-5) =0\)
k= \(\frac{10}{3}\) or \(\frac{5}{2}\)
If \(x+\frac{1}{x}= \frac{10}{3} = 3+\frac{1}{3}\)
x= 3 or \(\frac{1}{3}\)
\(x+\frac{1}{x }= \frac{5}{2} = 2+\frac{1}{2}\)
x = 2 or \(\frac{1}{2}\)
Bunuel
If \(6(x + \frac{1}{x})^2 − 35(x + \frac{1}{x}) + 50 = 0\), what could be the values of x
--> \((3(x+\frac{1}{x}) -10)(2(x+\frac{1}{x})-5)=0\)
\(3(x+\frac{1}{x})=10\) and \(2(x+\frac{1}{x})=5\)
\(x≠0\) --> \(3x^{2}-10x +3=0\) --> \((3x-1)(x-3)=0 \)
--> \(x=\frac{1}{3}\) or \(x= 3\)
\(x≠0\) --> \(2x^{2}-5x +2=0\) --> \((2x-1)(x-2)=0\)
--> \(x=\frac{1}{2}\) or \(x=2\)
The correct answer choice is E. (\(\frac{1}{2}\) and \(3\))
--> \((3(x+\frac{1}{x}) -10)(2(x+\frac{1}{x})-5)=0\)
\(3(x+\frac{1}{x})=10\) and \(2(x+\frac{1}{x})=5\)
\(x≠0\) --> \(3x^{2}-10x +3=0\) --> \((3x-1)(x-3)=0 \)
--> \(x=\frac{1}{3}\) or \(x= 3\)
\(x≠0\) --> \(2x^{2}-5x +2=0\) --> \((2x-1)(x-2)=0\)
--> \(x=\frac{1}{2}\) or \(x=2\)
The correct answer choice is E. (\(\frac{1}{2}\) and \(3\))
