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

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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, 10:49
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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, 19:54
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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, 13:50
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, 22:07
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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20 Mar 2017, 23:23
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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, 01:05
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, 14:03
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, 20:17
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]

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18 Jan 2018, 07:52
Bunuel 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

Factoring down the equation we have:

y(x^2 + 16x + 64) = 0

y(x + 8)(x + 8) = 0

y(x + 8)^2 = 0

Since y cannot be zero, x = -8, so we have 1 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 Jan 2018, 08:55
x^2y+16xy+64y=0 II Divide both sides by y
x^2+16x+64=0 II solve for x
x=-8

Would this approach be legitimate?

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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 Jan 2018, 10:02
Bunuel 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

It can be solved via b2-4ac=0, unique and 1 real root

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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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18 Jan 2018, 11:42
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.[/quote]

Hi Bunuel, if x^2y+16xy+64y=0

i something dont understand according rules of factoring quadratic

ab = 64
A+B= 16

so 8*8 = 64 (A could be 8)
8+8 =16 (B is 8)

hence X = 8 and not -8
so where from did you get negative -8 ?
pls explain thank you !

Hi chetan2u , perhaps you can help ? thanks!

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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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18 Jan 2018, 18:10
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Expert's post
dave13 wrote:
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.

Hi Bunuel, if x^2y+16xy+64y=0

i something dont understand according rules of factoring quadratic

ab = 64
A+B= 16

so 8*8 = 64 (A could be 8)
8+8 =16 (B is 8)

hence X = 8 and not -8
so where from did you get negative -8 ?
pls explain thank you !

Hi chetan2u , perhaps you can help ? thanks!

Hi...
You have gone wrong on calculating SUM of roots...
Product of roots = c/a=64=a*b.....
SUM of roots =-b/a=-16=a+b
So both a and b are -8
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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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19 Jan 2018, 06:09
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Hi...
You have gone wrong on calculating SUM of roots...
Product of roots = c/a=64=a*b.....
SUM of roots =-b/a=-16=a+b
So both a and b are -8

Hi chetan2u

ax^2 + bx + c = 0

Discriminant (D) = B^2-4Ac

Find roots:
X1 = -B+ sqrt D/ 2a

X2 = -B- sqrt D/2a

In our case, discriminant is equal to 0 (zero)

Hence:
x1 = -16+0/2 = - 8
X2 = -16-0/2 = -8
So both roots are equal to -8 ---> Correct?

Now lets solve the same problem through factorization

(x+a) (x+b)= 0

X^2+bx+ax+ab = 0

X^2+x (a+b)+ab = 0

Now lets get back to our case:

X^2y+16xy+64y = 0

Y (X^2+16+64) = 0

A is X^2
B is 16
C is 64

So 64 is our C term
The numbers that can multiply to make 64 are +8 and +8 (8*8 = 64)

16 is our B term

Now find the two factors of C that add up to your B term 8 and 8 Hence 8+8 = 16

Now plug in values I have chosen into factored equation (x+a) (x+b)= 0

(x+8) (x+8)= 0
Now solve for X by equating to 0
(x+8) =0 ---- > x = -8
(x+8)= 0 ----- >x = - 8

Is my understanding now correct ? So both roots are -8 (What if one eight is negative and other eight root is positive ? like this: -8 and +8 does it mean that equation has 2 roots ?

If my understanding is correct what role does Y play i our case ?

Thank you SO much ! Highly appreciated.

Last edited by dave13 on 19 Jan 2018, 10:17, edited 1 time in total.

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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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19 Jan 2018, 09:55
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Expert's post
dave13 wrote:
Hi...
You have gone wrong on calculating SUM of roots...
Product of roots = c/a=64=a*b.....
SUM of roots =-b/a=-16=a+b
So both a and b are -8

Hi chetan2u

ax^2 + bx + c = 0

Discriminant (D) = B^2-4Ac

Find roots:
X1 = -B+ sqrt D/ 2a

X2 = -B+ sqrt D/2a

In our case, discriminant is equal to 0 (zero)

Hence:
x1 = -16+0/2 = - 8
X2 = -16-0/2 = -8
So both roots are equal to -8 ---> Correct?

Now lets solve the same problem through factorization

(x+a) (x+b)= 0

X^2+bx+ax+ab = 0

X^2+x (a+b)+ab = 0

Now lets get back to our case:

X^2y+16xy+64y = 0

Y (X^2+16+64) = 0

A is X^2
B is 16
C is 64

So 64 is our C term
The numbers that can multiply to make 64 are +8 and +8 (8*8 = 64)

16 is our B term

Now find the two factors of C that add up to your B term 8 and 8 Hence 8+8 = 16

Now plug in values I have chosen into factored equation (x+a) (x+b)= 0

(x+8) (x+8)= 0
Now solve for X by equating to 0
(x+8) =0 ---- > x = -8
(x+8)= 0 ----- >x = - 8

Is my understanding now correct ? So both roots are -8 (What if one eight is negative and other eight root is positive ? like this: -8 and +8 does it mean that equation has 2 roots ?

If my understanding is correct what role does Y play i our case ?

Thank you SO much ! Highly appreciated.

highligh portion seems to have some error..
both are + signs.. must be a TYPING error as you have rectified it in the example later..

The understanding is perfect
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Re: How many real roots does the equation x^2y+16xy+64y=0 have if y < 0?   [#permalink] 19 Jan 2018, 09:55
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