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Bunuel
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


Solution


    • Let us assume that \(x+ \frac{1}{x }= t\) and substitute it in the given quadratic equation, we get
      o \(6t^2 - 35t + 50 = 0 ⟹ 6t^2 – 15t – 20t + 50 = 0 ⟹ 3t (2t - 5)-10(2t – 5) = 0 ⟹(2t - 5)(3t – 10) = 0\)
      o \(⟹ t = \frac{5}{2}\) or \(\frac{10}{3}\)
    • Since, we got two values of t, therefore there will be two cases,
      o Case 1: If \((x + \frac{1}{x}) = \frac{5}{2}\) then
         \((x + \frac{1}{x}) = \frac{(4+1)}{2} = 2 + \frac{1}{2}\)
         So, x can be \(2\) or \(\frac{1}{2}\)
      o Case 2: If \((x + \frac{1}{x}) = \frac{10}{3}\) then
         \((x + \frac{1}{x}) = \frac{(9+1)}{3 }= 3 + \frac{1}{3}\)
         So, x can be \(3\) or \(\frac{1}{3}\)
    • Thus, x can be \(2, \frac{1}{2 },3\) or \(\frac{1}{3}\)
Thus, the correct answer is Option E.
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See the attachment
Answer E
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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