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# How many real roots does the equation x^2y+16xy+64y=0 have if y < 0?

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Math Expert
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How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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28 Feb 2016, 11:49
Expert's post
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Difficulty:

65% (hard)

Question Stats:

54% (00:43) correct 46% (01:26) wrong based on 149 sessions

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How many real roots does the equation $$x^2y+16xy+64y=0$$ have if y < 0?

A. 0
B. 1
C. 2
D. 3
E. Infinite
[Reveal] Spoiler: OA

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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28 Feb 2016, 20:54
x^2y+16xy+64y=0
=> y ( x^2 + 16x + 64) = 0
=> y (x+8)^2 = 0

if y<0 , then x=-8
So although there are 2 factors , they are the same x=-8 .
The equations has 1 distinct real root .
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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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18 Mar 2017, 14:50
y(x^2+16x+64)=0
y=0
(x+8)=0
x=-8
hence B

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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20 Mar 2017, 23:07
If x= -8 then the equation equals 0, and the meaning of y doesn't change that. Therefore, there is an infinite number of solutions

I bet on E

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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21 Mar 2017, 00:23
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mill wrote:
How many real roots does the equation x^2y+16xy+64y=0 have if y < 0?

A. 0
B. 1
C. 2
D. 3
E. Infinite

If x= -8 then the equation equals 0, and the meaning of y doesn't change that. Therefore, there is an infinite number of solutions

I bet on E

You bet incorrectly. Please check the original post for the OA and the discussion above for solutions. The question asks for the number of values of x that satisfies the given equation if y < 0.

$$x^2y+16xy+64y=0$$

$$y(x+8)^2=0$$

Now, if y < 0 (so if $$y \neq 0$$), then the only way $$y(x+8)^2=0$$ to hold true is if x = -8.

Hope it's clear.
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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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21 Mar 2017, 02:05
x^2y+16xy+64y=0

or y(x+8)^2 = 0

since y is not equal to 0
therefore x+8 = 0
or x = -8

hence 1 solution.

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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17 Aug 2017, 15:03
ByjusGMATapp wrote:
x^2y+16xy+64y=0

or y(x+8)^2 = 0

since y is not equal to 0
therefore x+8 = 0
or x = -8

hence 1 solution.

So if Y could equal 0 then there would be 2 real roots?

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0? [#permalink]

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17 Aug 2017, 21:17
SamBoyle wrote:
ByjusGMATapp wrote:
x^2y+16xy+64y=0

or y(x+8)^2 = 0

since y is not equal to 0
therefore x+8 = 0
or x = -8

hence 1 solution.

So if Y could equal 0 then there would be 2 real roots?

Question mentioned y is less than 0

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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0?   [#permalink] 17 Aug 2017, 21:17
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