if x+y^2 = (x+y^2)^2 whats is the value of y? 1)x=y^2 : DS Archive
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# if x+y^2 = (x+y^2)^2 whats is the value of y? 1)x=y^2

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if x+y^2 = (x+y^2)^2 whats is the value of y? 1)x=y^2 [#permalink]

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31 Aug 2008, 18:27
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if x+y^2 = (x+y^2)^2 whats is the value of y?
1)x=y^2
2)xy^2=0

HELP NEEDED ,i got confused between equations
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31 Aug 2008, 19:18
i think A.

1) x=y^2
so x+y^2 = (x+y^2)^2 becomes 2y^2=4y^4. only 0 satisfies this. so y=0.

2) x +y^2=x^2+ y^4 + 2*x*y^2
or x +y^2=x^2+ y^4
insuff, as we dont know x.
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31 Aug 2008, 20:42
spriya wrote:
if x+y^2 = (x+y^2)^2 whats is the value of y?
1)x=y^2
2)xy^2=0

HELP NEEDED ,i got confused between equations

x+y^2 = (x+y^2)^2
x+y^2 = x^2+2xy^2+y^4

from1: x = y^2
x+y^2 = x^2+2xy^2+y^4
y^2+y^2 = y^4+2y^2y^2+y^4
2y^2 = 4y^4
y^2 = 2y^4
2y^4 - y^2 = 0
y^2 (2y^2 - 1) = 0
y = 0 or +/- 1/4 (but y = 1/4 doesnot satisfy the equation 2y^4 - y^2 = 0)
nsf...

from 2: xy^2 = 0
either x or y or both = 0
x+y^2 = x^2+2xy^2+y^4
x+y^2 = x^2+y^4

if x = 0:
y^2 = y^4
y^4 - y^2 = 0
y2 (y^2 -1) = 0
y = -1, 0, 1
so nsf..

from 1 and 2: y = 0.
C.
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Last edited by GMAT TIGER on 02 Sep 2008, 21:20, edited 1 time in total.
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01 Sep 2008, 18:10
GMAT TIGER wrote:
spriya wrote:
if x+y^2 = (x+y^2)^2 whats is the value of y?
1)x=y^2
2)xy^2=0

HELP NEEDED ,i got confused between equations

x+y^2 = (x+y^2)^2
x+y^2 = x^2+2xy^2+y^4

from1: x = y^2
x+y^2 = x^2+2xy^2+y^4
y^2+y^2 = y^4+2y^2y^2+y^4
2y^2 = 4y^4
y^2 = 2y^4
2y^4 - y^2 = 0
y^2 (2y^2 - 1) = 0
y = 0 or 1/2
nsf...

from 2: xy^2 = 0

either x or y or both = 0

x+y^2 = x^2+2xy^2+y^4
x+y^2 = x^2+y^4

if x = 0:
y^2 = y^4
y^4 - y^2 = 0
y2 (y^2 -1) = 0
y = -1, 0, 1

so nsf..

from 1 and 2: y = 0
C.

Oh great explanation !!!common solution of both ,i got wrong that part of the question.
OA is C indeed
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Its Now Or Never

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01 Sep 2008, 19:02
GMAT TIGER wrote:
spriya wrote:
if x+y^2 = (x+y^2)^2 whats is the value of y?
1)x=y^2
2)xy^2=0

HELP NEEDED ,i got confused between equations

x+y^2 = (x+y^2)^2
x+y^2 = x^2+2xy^2+y^4

from1: x = y^2
x+y^2 = x^2+2xy^2+y^4
y^2+y^2 = y^4+2y^2y^2+y^4
2y^2 = 4y^4
y^2 = 2y^4........A
2y^4 - y^2 = 0
y^2 (2y^2 - 1) = 0
y = 0 or 1/2.........B
nsf...

from 2: xy^2 = 0

either x or y or both = 0

x+y^2 = x^2+2xy^2+y^4
x+y^2 = x^2+y^4

if x = 0:
y^2 = y^4
y^4 - y^2 = 0
y2 (y^2 -1) = 0
y = -1, 0, 1

so nsf..

from 1 and 2: y = 0
C.

if we substitute B in A, it does not satisfy the eqn, so how 1/2 is a soln?
am i missing something?
Re: DS-equations   [#permalink] 01 Sep 2008, 19:02
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