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805+ (Hard)|   Algebra|      
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mathewmithun
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If x + y^2 = (x + y ^2)^2, what is the value of y? 03 Aug 2010, 12:31
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C
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95% (hard)

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47% (02:28) correct 53% (02:19) wrong based on 463 sessions

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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
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C
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Bunuel
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If x + y^2 = (x + y ^2)^2, what is the value of y? 03 Aug 2010, 12:56
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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) We have: \(x=y^2\) and \(xy^2=0\). Substitute \(x\) in the second equation: \(xy^2=y^2*y^2=y^4=0\) --> \(y=0\). Sufficient.

Answer: C.
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If x + y^2 = (x + y ^2)^2, what is the value of y? 07 Aug 2010, 10:53
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Bunuel
mitmat
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.

Answer: C.
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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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If x + y^2 = (x + y ^2)^2, what is the value of y? 07 Aug 2010, 11:16
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mainhoon
Bunuel
mitmat
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.

Answer: C.



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?
Show more

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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If x + y^2 = (x + y ^2)^2, what is the value of y? 05 Dec 2021, 02:30
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Stepping back and looking at the given equation we have:

One expression = (that same expression)^2

This can only be true for 2 real numbers: 0 or +1

Let: x + (y)^2 = U

(U) = (U)^2

(U)^2 - U = 0

(U) (U - 1) = 0

Either: U = 0 —-or—- U = +1

So

x + (y)^2 = 1

Or

x + (y)^2 = 0


S1: x = (y)^2


If y is not equal to 0 ——-> then no matter what the value of y is, X must be a positive value

If X is a positive value and (Y)^2 is a positive value, then

x + (y)^2 = 0 ——-> is not possible

So we could have this possibility if:

Case 1: X = 0 = (Y)^2 ——-> Y = 0


Case 2:

x + (y)^2 = 1

Since statement 1 tells us that (x) = (y)^2 —-> we can subsidies in (y)^2

(y)^2 + (y)^2 = 1

(y)^2 = (1/2)

Y = +sqrt(1/2) —— or ——- Y = (-) sqrt(1/2)

Y can take 3 different values


Statement 2

X (Y)^2 = 0

Zero product rule: either one of the terms or both of the terms must equal = 0

Case 1: x + (y)^2 = 0

Both terms can equal 0 ———> 1 possible answer is: Y = 0


Case 2: x + (y)^2 = 1

Y can equal 0 again and X would be = 1

Or

X = 0 ———> (y)^2 = 1

Y = + 1 ——or——- Y = (-)1

Y can be equal to 3 different values - not sufficient


Together:

S1: Y = 0 ——-or——- (Y)^2 = (1/2)

And

S2: Y = 0 ——- or ——— Y = (+)/(-) 1


Y = 0 is the only value that satisfies both statements

C - together sufficient to determine that Y = 0


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

(1) x = y ^2
(2) xy ^2 = 0
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If x + y^2 = (x + y ^2)^2, what is the value of y? 24 May 2022, 09:26
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