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# roots of equation

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Intern
Joined: 09 Nov 2009
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roots of equation [#permalink]

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11 Nov 2009, 09:22
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The equation y-(5/y-3)=3-(5/y-3) has:

A) an infinite number of roots, all of them real
B) exactly two roots, both nonintegral
C) exactly two roots, both integral
D) exactly one root, which is integral
E) no roots

I get 3 for a root (D), but ans is E?
VP
Joined: 05 Mar 2008
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Re: roots of equation [#permalink]

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11 Nov 2009, 09:34
The equation y-(5/y-3)=3-(5/y-3) has:

A) an infinite number of roots, all of them real
B) exactly two roots, both nonintegral
C) exactly two roots, both integral
D) exactly one root, which is integral
E) no roots

I get 3 for a root (D), but ans is E?

3 can't be a root because if you substitute 3 for y you get undefined because the denominator is (y-3)
Manager
Joined: 23 Jun 2009
Posts: 155
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Re: roots of equation [#permalink]

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11 Nov 2009, 15:46
lagomez wrote:
The equation y-(5/y-3)=3-(5/y-3) has:

A) an infinite number of roots, all of them real
B) exactly two roots, both nonintegral
C) exactly two roots, both integral
D) exactly one root, which is integral
E) no roots

I get 3 for a root (D), but ans is E?

3 can't be a root because if you substitute 3 for y you get undefined because the denominator is (y-3)

Yes that is precisely why the equation doesn't have any roots. Hence, answer is
"E"
Manager
Joined: 10 Sep 2009
Posts: 119
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Re: roots of equation [#permalink]

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12 Nov 2009, 03:18
Could someone confirm OA please?
Re: roots of equation   [#permalink] 12 Nov 2009, 03:18
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# roots of equation

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