What is the product of all the solutions of (x + 2)^2 = |x + 2|?

Question

A. -6
B. -2
C. 2
D. 6
E. 12

JeffTargetTestPrep · Accepted Answer

(x + 2)^2 = |x + 2|

|x + 2|^2 = |x + 2|

If the square of an expression is equal to the expression itself, then that expression must be equal to 1 or 0.

If |x + 2| = 1, then x = -1 OR x = -3.

If |x + 2| = 0, then x = -2.

So, the product of all the solutions is:

(-1)(-3)(-2) = -6

Answer: A

Note: In the solution, we used the rule that |y|^2 = y^2 for all y.

gmatophobia · Answer

y = x + 2

(y)^2 = |y|

squaring both sides

y^4 - y^2 = 0

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

Case 1:  y^2 = 0 
   y = 0 
  x + 2 = 0 
 x = -2     
 Case 2 :  y^2 - 1 = 0 
   y = +1 
  x + 2 = +1 
 x = -1   
  y = -1 
  x + 2 = -1 
 x = -3

Net product =  -2 * -3 * -1 = -6

Option A

raianshul · Answer

(x + 2)^2 = |x + 2|

case1: when x>=-2 
(x + 2)^2 = (x + 2)
(x+2)*(x+2-1)=0
(x+2)*(x+1)=0
x=-2, -1

case2: when x<-2 
(x + 2)^2 + (x + 2) =0
(x+2)*(x+3)=0
x=-3, x=-2
x cannot be equal to -2, since x<-2, hence x=-3 is the only root from case2

So roots are x=-1,-2 and -3
 Product = -6, hence A should be the right choice.

Oppenheimer1945 · Answer

visualizing plot of  |x|=x^2 , solutions are  x=-1,0,1

The function given in question is shifted left by  2  units, The new solutions will be  x=-1-2,0-2,1-2=-3,-2,-1

Product = -6

Archit3110 · Answer

(x + 2)^2 = |x + 2|

x^2+4x+4 = lx+2l
we get values of x = -1,-2,-3
product is -6
option A

madboy1105 · Answer

Why can't we cancel (x+2)^4 = (x+2)^2 to make (x+2)^2 = 1?

gmatophobia · Answer

madboy1105  - We can't divide on both sides of the equation by  (x+2)^2  and arrive at  (x +2)^2 = 1  as the value of  (x+2)^2  can be zero (at x = -2).

Note: Whenever one cancels out value by dividing both sides by a common term,  (x+2)^2  in this example, one needs to ensure that the divisor is not zero. Division by zero is an invalid operation.

As we cannot rule out that possibility in this case, the division operation is not valid and we cannot cancel out terms as you're expecting.

Hope this helps.

adewale223 · Answer

(x + 2 )^2 = /x + 2 /

x^2 + 4x + 4 = /x + 2 /

Case 1

When x + 2 is positive.

x^2 + 4x + 4 = x + 2

x^2 + 4x + 4 -x - 2 = 0

x^2 + 3x + 2 = 0

x = -2
x = -1

Case 2

When x + 2 is negative.

x^2 + 4x + 4 = -x - 2

x^2 + 5x + 6 = 0

x = -2

x = -3

Substituting the values shows x could be -1, -2, -3

Sum = -6

Answer choice A

ruis · Answer

Shouldn´t case 2 factor to (x+1) (x+6)?? Hence x= -1 and x= -6. Their product is 6 and their difference is 5

Bunuel · Answer

When you expand (x + 1)(x + 6) do you get x^2 + 5x + 6? No. You get x^2 + 7x + 6. Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm Hope it helps.

einstein801 · Answer

Hi  Bunuel , I followed this method and ended up with the same results but did -2*-1*-3*-2.

Question did not state distinct answers so 12 could be the answer as well.

Bunuel · Answer

Firstly, the solutions of the equation are -1, -2, and -3. No one would say that the solutions are -1, -3, -2, AND -2 AGAIN.

Also, the method above lacks details to be 100% correct.

In case 1, when x + 2 is positive, it means the range is x > -2. So, you should discard x = -2, which you get for that case because it is not in the range.

In case 2, we should consider the scenario when x + 2 ≤ 0, giving the range x ≤ -2. Both x = -2 and x = -3 are in the range, so both are acceptable.

Thus, if done properly, that method gives only three roots: -1, -2, and -3.

Hope it helps.

FlorianSchltr · Answer

A algebraic way of calculating cases:

|x + 2| is the equivalent of ((x+2)^2)^1/2, so basically the square root of a squared number is equal to the absolute value

(x+2)^2 = ((x+2)^2)^1/2 => square
(x+2)^4 = (x+2)^2 => subtract - (x+2)^2
(x+2)^4 - (x+2)^2 =0 => Factorize (x+2)^2 (You only need to factorize and not solve for difference in squares which you could also use here if you look closely cause you want the equation to be in a factorized form = 0 to use the property that one factor has to be 0 then
(x+2)^2 * (((x+2)^2-1)) = 0; Therefore (x+2)^2 = 0 or (x+2)^2-1 = 0
(x+2)^2 = 0 if x = -2
(x+2)^2-1 = 0 if x = -3 or x = -1
Solution = -1*-3*-2 = -6

CapnSal · Answer

Thanks  Bunuel , does this mean the "distinct" implication is implied ?

Bunuel · Answer

The point is, if you solve the question correctly as demonstrated in many of the above posts, you don't even get x = -2 twice. Therefore, it doesn't even need to be implied.

Kinshook · Answer

Asked: What is the product of all the solutions of (x + 2)^2 = |x + 2|?

(x+2)ˆ2 = |x+2|
|x+2|ˆ2 - |x+2| = 0
|x+2| ( |x+2| - 1) = 0
Case 1: |x+2| = 0
x = -2
Case 2: |x+2| - 1 =0
|x+2| = 1
x = -1 or -3

x = {-3,-2,-1}

The product of all the solutions = (-3)(-2)(-1) = -6

IMO A

DanTheGMATMan · Answer

Prethink what values on each side will yield equal results:

https://www.youtube.com/watch?v=UHvrMTc9s0g

sanjanagupta27 · Answer

(x+2)²=|x+2|
Squaring both sides 
(X+2)⁴=(x+2)²
Let (x+2)²=t
t²-t=0
t(t-1)=0
t=0,t=1
(X+2)²=0,x=-2
(X+2)²=1,x=+-1-2
X=-1 and -3
Product=-6 ans

Posted from my mobile device

GMATinsight · Answer

Let, x+2 = a

then (x + 2)^2 = |x + 2| becomes  a^2 = |a|  which is possible for

a = ±1 OR a = 0
i.e x+2 = ±1 OR x+2 = 0

if x+2 = 1 then x = -1
if x+2 = -1 then x = -3
if x+2 = 0 then x = -2

Product of all possible values of x = (-1)*(-3)*(-2) = -6

Answer: Option A

What is the product of all the solutions of (x + 2)^2 = |x + 2|?

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What is the product of all the solutions of (x + 2)^2 = |x + 2|?

Prep Toolkit

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