Is x^2+2x+1=y^2+2y+1?

Question

Is  x^2+2x+1=y^2+2y+1 ?
(1)  x
eq{y} 
(2) x>y>0

Chetan's  questions 
kudos to correct answer

Akash720 · Answer

+1 for D 
Given:
 x^2+2x+1=y^2+2y+1 ? 
 (x+1)^2=(y+1)^2 ? 
 x+1=y+1 ? 
 x=y ?

We need to find if x=y.

Statement 1:
 x
eq{y} 
 Hence sufficient 
Statement 2:
x>y>0
 Hence Sufficient

FillFM · Answer

(1)  x ≠ y

let: x = 3 and y = 4

so,  x^2 + 2x + 1 ≠ y^2 + 2y + 1

now, let x = 0 and y = -2

so,  x^2 + 2x + 1 = y^2 + 2y + 1

Insufficient.

(2) x>y>0

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

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

But, x>y>0

So,  x^2 + 2x + 1 ≠ y^2 + 2y + 1

Sufficient.

Option B

PKN · Answer

Rephrasing question stem:- Is (x+1)^2=(y+1)^2 ? The statement that says either (x+1)^2 eq{(y+1)^2} or (x+1)^2={(y+1)^2} would be sufficient. St1:- x eq{y} Here we have three cases: a) If x>y, then (x+1)^2<(y+1)^2 (say x=-2 and y=-3) b) If x>y, then (x+1)^2>(y+1)^2 (say x=3 and y=2) c) If x>y, then (x+1)^2=(y+1)^2 (say x=0, y=-2) d)If x(y+1)^2 (say x=-3 and y=-2) e)If xy>0 or x and y are positive numbers with x>y. So, (x+1)^2>(y+1)^2 will hold true. Sufficient. Ans. (B)

Princ · Answer

OA: B

Is  x^2+2x+1=y^2+2y+1 ?
reducing the question  x^2+2x+1-y^2-2y-1=0 
 x^2-y^2+ 2x-2y=0 
 (x-y)(x+y)+2(x-y)=0 
Question becomes  Is  (x-y)(x+y+2)=0 ?

Statement 1 :  x
eq{y} 
As  {x-y}
eq{0} , Questions become  Is   (x+y+2)=0 ?
Now plugging in values ,

First taking  x=-2,y=0

L.H.S   : (x+y+2)=-2+0+2=0 
 R.H.S   =0 
 L.H.S  =  R.H.S 
Answer to  Is   (x+y+2)=0 ? becomes yes

Now taking  x=2,y=3

L.H.S   : (x+y+2) = 2+3+2 =7 
 R.H.S  = 0 
 {L.H.S}
eq{R.H.S} 
Answer to  Is   (x+y+2)=0 ? becomes No

Statement 1 alone is not sufficient.

Statement 2 :  x>y>0

This means  (x-y)  =some +ve number,  x+y+2  will also be some +ve number.
So  (x-y)(x+y+2)>0 
Answer to question  Is   (x-y)(x+y+2)=0 ? is always No
Statement 2 alone is sufficient.

Hero8888 · Answer

Is  x^2+2x+1=y^2+2y+1  ,  (x+1)^2=(y+1)^2  ,  (x+1)^2-(y+1)^2=0  ,  (x+1-y-1)(x+1+y+1)=0  ,  (x-y)(x+y+2)=0 , x=y or x+y=-2

(1)  x
eq{y} , but we don't know whether x+y=-2 - Not Suff
(2) x>y>0 - means that  x
eq{y}  and  x+y>0  - Suff

Answer B.

Mo2men · Answer

x^2+2x+1=y^2+2y+1 
 (x+1)^2=(y+1)^2 
 |x + 1|=|y + 1| 
 
2 cases to open modulus

x+1=y+1 ..........x=y

x+1=-y-1 .........x+y=-2

The question becomes:

x=y or x+y=-2

(1)  x
eq{y}

It could be that x + y =-2

Let x =1 & y =-3............Answer is Yes

or

Any any other values

Let x =1 & y =-2............Answer is No

Insufficient

(2) x>y>0

It is clear that  x
eq{y}  and psotive numbers can't produce negative number (-2). Therefore it is certain answer with NO.

Answer: B

dave13 · Answer

hey PKN can you please explain how from here x^2+2x+1=y^2+2y+1 you got this (x+1)^2=(y+1)^2 ? thank you

PKN · Answer

I hope you know the basic algebra formula:-

(a+b)^2=a^2+2ab+b^2

Similarly,  (x+1)^2=x^2+2*x*1+1^2=x^2+2x+1

Hope it helps.

ccooley · Answer

It's (B).

Be careful when you go from here:

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

to here:

x+1=y+1 ?

It actually isn't quite that simple. For instance, if  a^2 = b^2 , that doesn't mean that  a = b . a could be -2 and b could be positive 2.

It actually simplifies like this:

|x + 1| = |y + 1| ?

Which can be true in two situations: if x and y are equal, or if one or both of them is negative (for instance, x = -1.5 and y = -0.5, or x = -10 and y = 8.)

If they're both positive, and they aren't equal, you know the answer to the question is 'no'.

Raksat · Answer

we can simplify the question as
|x+1|=|y+1|
Statemment1
consider x=-2 and y =0, then we have |x+1|=|y+1| = 1
Consider x=1 and y =2 , then we have |x+1|≠|y+1|
Hence insufficient
Statement 2
X>Y>0, we can deduce that both are positive. And hence we can never have |x+1|=|y+1| for any values of x and y
Hence sufficient
Therefore IMO the answer should be B

dave13 · Answer

hi Mo2men can you please explain how you opened modulus in second case |x + 1|=|y + 1| x+1=-y-1 .........x+y=-2 i thought we need to open moduls on both sides like this -x-1=-y-1 why it is not correct opening ?

pushpitkc maybe you can help with this

thank you

Mo2men · Answer

Hi Dave,

The modulus lesson is one of the important lessons that you need to take it step by step to learn sufficiently. Loot at  chetan2u  signuatre as he has created nice lesson in absolute value besides GC math book or any other source you like.

I will try gladly to help

When You have both side with absolute value:

1- You will fix one side and change the alter the sign of other side. and therefore,

2- you must consider two cases

Let's apply in the above:

|x + 1|=|y + 1|

Case 1:

1- Fix the the LHS and 
2- Fix the sign of RHS to be positive which means it will be as it is.

x + 1 = y + 1............X = Y

Case 2:

1- Fix the the LHS and 
2- Change the sign of RHS to be Negative.

x + 1 = -(y + 1)............x + 1 = -y -1..............x + y = -2

What you did wrong is the you multiplied the whole equation in negative sign.

I hope it hlpes

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Is x^2+2x+1=y^2+2y+1?

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