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# If x + y^2 = (x + y ^2)^2, what is the value of y?

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Joined: 22 Jul 2010
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If x + y^2 = (x + y ^2)^2, what is the value of y?  [#permalink]

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03 Aug 2010, 12:31
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If x + y^2 = (x + y ^2)^2, what is the value of y?

(1) x = y ^2
(2) xy ^2 = 0
Math Expert
Joined: 02 Sep 2009
Posts: 56371

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03 Aug 2010, 12:56
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mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

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07 Aug 2010, 10:53
Bunuel wrote:
mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

Bunuel
For Point (2) above, xy2=0, we examine the case where x=0 to find values of y, what about when y=0, which means x-x^2=0 =>
x(x-1)=0; x=0 or x=1
Why did we not examine this sequence?
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07 Aug 2010, 11:16
mainhoon wrote:
Bunuel wrote:
mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

Bunuel
For Point (2) above, xy2=0, we examine the case where x=0 to find values of y, what about when y=0, which means x-x^2=0 =>
x(x-1)=0; x=0 or x=1
Why did we not examine this sequence?

The question is "what is the value of $$y$$". So, we don't care what $$x$$ is.

Statement (2) is: $$xy^2=0$$. This statement would be sufficient if it could give us the single numerical value of $$y$$.

From (2):
Either $$y=0$$ OR $$x=0$$. $$y=0$$ is obvious solution and we should concentrate on another option, which is $$x=0$$, to see whether this case can give the same or different solutions of $$y$$ and thus to determine whether this statement is sufficient.

$$x=0$$ gives 3 values of $$y$$: 0, 1, and -1, so we can say that this statement is not sufficient.

Hope it's clear.
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07 Aug 2010, 11:56
So for (2) itself, y=0; x=0 or 1 is a valid solution. With x=0, we end up with y=0,1,-1 and thus if we went for (1)+(2), we can see that x=0,y=0 is the only solution that is common across all cases. In this case it so happens that the obvious solution (0,0) is what we end up with, however, in the general case (say for another problem), would you still not check the y=0 solutions for x, because the questions asks for the value of not x? I wanted to know the thought process. Thanks

Another question - when quote an answer in the forum, I see the window flicker and move up and down randomly (I have to try hard to keep the window and the space I am typing in in view.. Anything I am doing wrong here? Thanks
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07 Aug 2010, 12:08
mainhoon wrote:
So for (2) itself, y=0; x=0 or 1 is a valid solution. With x=0, we end up with y=0,1,-1 and thus if we went for (1)+(2), we can see that x=0,y=0 is the only solution that is common across all cases. In this case it so happens that the obvious solution (0,0) is what we end up with, however, in the general case (say for another problem), would you still not check the y=0 solutions for x, because the questions asks for the value of not x? I wanted to know the thought process. Thanks

Another question - when quote an answer in the forum, I see the window flicker and move up and down randomly (I have to try hard to keep the window and the space I am typing in in view.. Anything I am doing wrong here? Thanks

If I'm asked to find y and say I get that y=some value, then why should I find the value of another unknown?

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If x + y^2 = (x + y^2)^2, what is the value of y?  [#permalink]

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21 Jan 2014, 10:35
If x + y^2 = (x + y^2)^2, what is the value of y?

(1) x = y^2
(2) xy^2 = 0

OE
(1): Substitute y^2 for x into x + y^2 = (x + y^2)^2 → y^2 + y^2 = (y^2 + y^2)^2 → 2y^2 = 4y^4
As y may be equal to 0, cannot divide by y^2
2y^2 = 4y^4 → 4y^4 – 2y^2 = 0
Factor out 2y2 from left side of equation to give 2y^2(2y^2 – 1) = 0
→ either 2y^2 = 0 or 2y^2 – 1 = 0.
When product of a group of terms is 0, at least one of the terms must be 0
If 2y^2 = 0 → y = 0
If (2y^2 – 1) = 0 → 2y^2 = 1 → y = ±√(1/2)
→ y = 0 or ±√(1/2)
Insufficient

(2): Either x = 0 or y = 0
If x = 0, substitute 0 for x into x + y^2 = (x + y^2)^2
→ 0 + y^2 = (0 + y^2)^2 → y^2 = (y^2)^2 → y^2 = y^4 → y^4 – y^2 = 0 → y^2(y^2 – 1) = 0
If y^2 = 0 → y = 0
If y^2 – 1 = 0 → y^2 = 1 & y = ±1
If x = 0, → y = –1, 0, or 1
Insufficient

Combined: From (1) y = 0 or ±√(1/2). From (2) y = –1, 0, or 1 → y = 0
Sufficient

Hi, I want to know if we have more simple way to solve this question, please.
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Joined: 02 Sep 2009
Posts: 56371
Re: If x + y^2 = (x + y^2)^2, what is the value of y?  [#permalink]

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21 Jan 2014, 10:44
1
goodyear2013 wrote:
If x + y^2 = (x + y^2)^2, what is the value of y?

(1) x = y^2
(2) xy^2 = 0

OE
(1): Substitute y^2 for x into x + y^2 = (x + y^2)^2 → y^2 + y^2 = (y^2 + y^2)^2 → 2y^2 = 4y^4
As y may be equal to 0, cannot divide by y^2
2y^2 = 4y^4 → 4y^4 – 2y^2 = 0
Factor out 2y2 from left side of equation to give 2y^2(2y^2 – 1) = 0
→ either 2y^2 = 0 or 2y^2 – 1 = 0.
When product of a group of terms is 0, at least one of the terms must be 0
If 2y^2 = 0 → y = 0
If (2y^2 – 1) = 0 → 2y^2 = 1 → y = ±√(1/2)
→ y = 0 or ±√(1/2)
Insufficient

(2): Either x = 0 or y = 0
If x = 0, substitute 0 for x into x + y^2 = (x + y^2)^2
→ 0 + y^2 = (0 + y^2)^2 → y^2 = (y^2)^2 → y^2 = y^4 → y^4 – y^2 = 0 → y^2(y^2 – 1) = 0
If y^2 = 0 → y = 0
If y^2 – 1 = 0 → y^2 = 1 & y = ±1
If x = 0, → y = –1, 0, or 1
Insufficient

Combined: From (1) y = 0 or ±√(1/2). From (2) y = –1, 0, or 1 → y = 0
Sufficient

Hi, I want to know if we have more simple way to solve this question, please.

Merging similar topics. Please refer to the solution above.
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Re: If x + y^2 = (x + y ^2)^2, what is the value of y?  [#permalink]

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07 Aug 2018, 06:35
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Re: If x + y^2 = (x + y ^2)^2, what is the value of y?   [#permalink] 07 Aug 2018, 06:35
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