What is the sum of all the solutions of |y + 1| = (y + 1)^2?

Question

A. 1
B. 0
C. -1
D. -2
E. -3

gmatophobia · Accepted Answer

y + 1 = x

|x| = (x)^2

Squaring on both sides we get

x^2 = x^4

x^2(1 - x^2) = 0

Possible solutions

Case 1 :  x = 0 
 
  Case 2 :  x^2 = 1

Case 1 : x = 0

y + 1 = 0

y = -1

Case 2 :  x^2 = 1

(y + 1)^2 = 1

y^2 + 2y + 1 = 1

y(y + 2) = 0

y = 0; y = -2

Sum of all solutions = -1 + 0 + -2 = -3

Option E

BrushMyQuant · Answer

We need to find What is the sum of all the solutions of |y + 1| = (y + 1)^2?

We have two cases as we have |y+1| 
 
   -   Case 1:  y+1 ≥ 0 or y ≥ -1
 
=> | y + 1 | = y + 1
=> y + 1 =  (y+1)^2 
=>  (y+1)^2  - (y+1) = 0
=> (y+1)*(y + 1 - 1) = 0
=> (y+1)*y = 0
=> y = 0, -1

But the condition was y ≥ -1 and y = 0 and -1 are both ≥ -1
  => y = 0, -1 are  SOLUTIONS   
  -   Case 2:  y + 1 ≤ 0 or y ≤ -1
 
 => | y + 1 | = -(y + 1)
=> -(y + 1) =  (y+1)^2 
=>  (y+1)^2  + (y+1) = 0
=> (y+1)*(y + 1 + 1) = 0
=> (y+1)*(y + 2)= 0
=> y = -1, -2

But the condition was y ≤ -1 and y = -1 and -2 are both ≤ -1
  => y = -1, -2 are  SOLUTIONS

=> Sum of all the solutions for y = -2 + -1 + 0 = -3

So,  Answer will be E 
Hope it helps!

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https://www.youtube.com/watch?v=huFiBk8PaGY

Bunuel · Answer

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

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

|y + 1| - (|y + 1|)^2 = 0

|y + 1|(1 - |y + 1|) = 0

|y + 1| = 0  or  |y + 1| = 1

|y + 1| = 0  gives  y = -1 .

|y + 1| = 1  gives  y = 0  or  y = -2 .

Therefore, the sum of the roots is  -1 + 0 + (-2) = -3 .

Answer: E

saynchalk · Answer

gmatophobia 
 Bunuel

Thank you for the quick response and for the extensive details.

Oppenheimer1945 · Answer

put  y+1=t ,

|t|=t^2  gives  t=-1,0,1  (graphically)

=> y=-1,-2,0. 
Sum of values=  -3

Rave25 · Answer

I solved this question in this way as shown in attached file. Please let me know what i did wrong?

Posted from my mobile device

BrushMyQuant · Answer

|x| = |x| * |x|
=> |x| * |x| - |x| = 0
=> |x| * (|x| - 1) = 0
=> |x| = 0, 1

You cancelled out |x| on both the sides, instead of that you have to consider one solution as |x| = 0

Hope it helps!

mbaaccount1234 · Answer

Hey,
Why doesnt it work to use (y+1) = (y + 1)^2 and -(y+1) = (y + 1)^2 ?
Tks!

Unknown5612 · Answer

Two cases:

1) y>1:

y+1=(y+1)^2
y^2+y=0
solve, will get y=0,-1

2) y<-1
-y-1=(y+1)^2
y^2+3y+2=0
solve, y=-2,-1

sum of roots=0+-1--3

eega · Answer

Given |y+1| = (y+1)^2
For y<-1
Opening modulus
-y-1=y^2 + 2y +1

Solving this we get y = -1 or -2, we can take only -2 (as y<-1)

For y>= -1
Opening modulus
y+1 = y^2 + 2y +1
solving we get y = 0 or y= -1

Hence adding all we get y = -3  Option E

siddhantvarma · Answer

It works; just solve the two equations separately and add all the values you get as y for the two equations. You'll get 0, -1 and -2. You'll get -1 twice so remember to only consider it once.

Ilanchezhiyan · Answer

could you explain why we consider it once? I didnt get that part. Thanks

Bunuel · Answer

We consider -1 only once because we’re counting distinct solutions. Even if it comes from two different equations, it’s still the same value.

What is the sum of all the solutions of |y + 1| = (y + 1)^2?

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We have two cases as we have \|y+1\|
-Case 1: y+1 ≥ 0 or y ≥ -1 => \| y + 1 \| = y + 1 => y + 1 = \((y+1)^2\) => \((y+1)^2\) - (y+1) = 0 => (y+1)(y + 1 - 1) = 0 => (y+1)y = 0 => y = 0, -1 But the condition was y ≥ -1 and y = 0 and -1 are both ≥ -1 => y = 0, -1 are SOLUTIONS	-Case 2: y + 1 ≤ 0 or y ≤ -1 => \| y + 1 \| = -(y + 1) => -(y + 1) = \((y+1)^2\) => \((y+1)^2\) + (y+1) = 0 => (y+1)(y + 1 + 1) = 0 => (y+1)(y + 2)= 0 => y = -1, -2 But the condition was y ≤ -1 and y = -1 and -2 are both ≤ -1 => y = -1, -2 are SOLUTIONS

GMAT Problem Solving (PS) Questions

What is the sum of all the solutions of |y + 1| = (y + 1)^2?

Prep Toolkit

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More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards