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What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?

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Senior Manager
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What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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New post 26 Mar 2019, 17:34
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

75% (01:41) correct 25% (01:53) wrong based on 20 sessions

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What is the value of \(y\) if \(y > 0\) and \(4 - 3y(y-\frac{4}{y}) = 41 - 4y^2?\)

A) 25
B) 20
C) 12
D) 8
E) 5
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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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New post 26 Mar 2019, 18:24
I am probably getting my algebra totally wrong, but I don't see how any of those options can be a correct answer, can you see whether there is a typo?
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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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New post 26 Mar 2019, 18:42
By simplifying ,
the equation becomes 4y^2 -3y -25 = 0
This equation will have imaginary roots!!! Is there something wrong in the question

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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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New post 26 Mar 2019, 19:34
The question has a typo. The denominator y is just below 4/y, not below entire (y-4)/y
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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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New post 26 Mar 2019, 20:26
Hi guys,
The problem should be 4-3y(y-(4/y))=41-4y^2

Simplifying LHS we get
4-3y(y^2-4)/y
I.e 4-3y^2+12=41-4y^2
Substraction yields
y^2=25.
Therefore answer is y=5 as y>0
So answer is E

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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?   [#permalink] 26 Mar 2019, 20:26
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