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If z ≠ 1 and z^2 − 2z + 20/(z −1) =20, then how many negative values

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If z ≠ 1 and z^2 − 2z + 20/(z −1) =20, then how many negative values  [#permalink]

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New post Updated on: 31 May 2017, 23:01
5
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

53% (02:18) correct 47% (02:19) wrong based on 57 sessions

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If z ≠ 1 and \(z^2−2z+\frac{20}{z−1}=20\), then how many negative values can z take ?

A. None
B. One
C. Two
D. Three
E. A finite number greater than three

Originally posted by niteshwaghray on 31 May 2017, 22:57.
Last edited by Bunuel on 31 May 2017, 23:01, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If z ≠ 1 and z^2 − 2z + 20/(z −1) =20, then how many negative values  [#permalink]

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New post 31 May 2017, 23:21
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1
niteshwaghray wrote:
If z ≠ 1 and \(z^2−2z+\frac{20}{z−1}=20\), then how many negative values can z take ?

A. None
B. One
C. Two
D. Three
E. A finite number greater than three



\(z^2−2z+\frac{20}{z−1}=20\)

\(z^2*(z - 1) - 2z*(z - 1) + 20 = 20*(z - 1)\)

\(z^3 - 3z^2 - 18z + 40 = 0\)

\(z = 2\)

\((z - 2)*(z^2 - z - 20) = 0\)

\((z - 2)*(z - 5)*(z + 4) = 0\)

\(z = 2, 5, -4\)

So z can take one negative value.

Answer (B)

For more on this, check: https://www.veritasprep.com/blog/2013/0 ... rd-degree/
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Re: If z ≠ 1 and z^2 − 2z + 20/(z −1) =20, then how many negative values  [#permalink]

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New post 22 Mar 2019, 13:01
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Re: If z ≠ 1 and z^2 − 2z + 20/(z −1) =20, then how many negative values   [#permalink] 22 Mar 2019, 13:01
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