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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 22 Jun 2024, 03:36
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
Show more
­
Keep in mind:

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

Hence we can write the given equation as:\(|x+2|^2 - |x+2| = 0 \)

\(|x+2| * ( |x+2| - 1) = 0\)

So either |x+2| = 0 which means x = -2
or
|x+2| - 1 = 0 which means x is -1 or -3

Product of all solutions = -1 * -2 * -3 = -6

Answer (A)­

Video on solving absolute values:
https://youtu.be/oqVfKQBcnrs
 
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? Updated on: 01 Apr 2025, 19:29
Originally posted by HarshavardhanR on 04 Aug 2024, 00:52.
Last edited by HarshavardhanR on 01 Apr 2025, 19:29, edited 1 time in total.
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 04 Apr 2025, 03:43
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

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

x+2 can be ±1 gives us two possible solutions
x = -1 and -3


Another solution that can be seen with just trial and error = -2 so that we have 0 on both sides

i.e. Product of all solutions = (-1)*(-3)*(-2) = -6

Answer: Option A

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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 24 Jul 2025, 22:21
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
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Why aren't we considering x=0 as one of the solutions. Since, it also satisfies the equation.
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 25 Jul 2025, 00:11
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
Why aren't we considering x=0 as one of the solutions. Since, it also satisfies the equation.
Show more

0 does not satisfy the equation. For x = 0:

(x + 2)^2 = 4

|x + 2| = 2
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 27 Sep 2025, 10:15
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Can you help me and explain why this approach is incorrect?

(x+2)^2 = |x+2|
Since modulus and square is always positive,
(x+2)^2 - (x+2)=0
(x+2)(x+2-1)=0
(x+2)(x+1)=0
x=-2, -1
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 27 Sep 2025, 11:46
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IJIJ
Can you help me and explain why this approach is incorrect?

(x+2)^2 = |x+2|
Since modulus and square is always positive,
(x+2)^2 - (x+2)=0
(x+2)(x+2-1)=0
(x+2)(x+1)=0
x=-2, -1
Show more
The mistake is that you replaced |x+2| with (x+2). That only holds if x+2 ≥ 0. For x+2 < 0, |x+2| = -(x+2), which you ignored. Please read carefully and study the previous two pages of the discussion.
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 28 Sep 2025, 22:33
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IJIJ
Can you help me and explain why this approach is incorrect?

(x+2)^2 = |x+2|
Since modulus and square is always positive,
(x+2)^2 - (x+2)=0
(x+2)(x+2-1)=0
(x+2)(x+1)=0
x=-2, -1
Show more
IJIJ Your statement "(x+2)2 - (x+2) = 0" assumes that \(|x+2| = (x+2)\) for all values of \(x\). This is the key misconception!

Why This Doesn't Work:

While both \((x+2)^2\) and \(|x+2|\) are indeed non-negative, they are different functions:
- \(|x+2| = (x+2)\) only when \(x+2 \geq 0\)
- \(|x+2| = -(x+2)\) when \(x+2 < 0\)

The Correct Approach:

We must consider both cases for the absolute value:

Case 1: When \(x+2 \geq 0\) (i.e., \(x \geq -2\))
\(|x+2| = (x+2)\)
So: \((x+2)^2 = (x+2)\)
\((x+2)^2 - (x+2) = 0\)
\((x+2)(x+2-1) = 0\)
\((x+2)(x+1) = 0\)
This gives \(x = -2\) or \(x = -1\)
Both satisfy \(x \geq -2\), so both are valid! ✓

Case 2: When \(x+2 < 0\) (i.e., \(x < -2\))
\(|x+2| = -(x+2)\)
So: \((x+2)^2 = -(x+2)\)
\((x+2)^2 + (x+2) = 0\)
\((x+2)(x+2+1) = 0\)
\((x+2)(x+3) = 0\)
This gives \(x = -2\) or \(x = -3\)
Only \(x = -3\) satisfies \(x < -2\), so it's valid! ✓

Solutions: \(x = -3, -2, -1\)
Product: \((-3) \times (-2) \times (-1) = -6\)
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 06 Jun 2026, 07:07
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Quote:
|x + 2|^2 = |x + 2|
Show more
At this stage of the solution, given side will be positive, we can cancel out |x+2|


Thus giving |x+2|=1
Case 1 : x+2>0

Thus x+2=1 thus x=-1
Check does this give x+2>0 when x = -1 : Yes this satisfied

Caution : dont do given we need equation to be positive but x=-1 is negative thus invalid solution, what needs to be checked is x+2>0 and not x>0


∣x+2∣ is nonnegative. For the case ∣x+2∣ ≠ 0, we can divide both sides by ∣x+2∣, giving ∣x+2∣ = 1.
Also consider the case ∣x+2∣ = 0 separately, since dividing by ∣x+2∣ would eliminate that solution.


JeffTargetTestPrep


(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.
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 26 Jul 2026, 05:02
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A^2= |A| means A=0 or A=1 or A=-1 now A=X+2

X+2= -1,0,1 solution X= -3, -2, -1 product is -6
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 29 Jul 2026, 05:01
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here shouldnt there be another case which is when it is 0?
adewale223
(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
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 29 Jul 2026, 06:16
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ananya88888888888
here shouldnt there be another case which is when it is 0?

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0 does not satisfy (x + 2)^2 = |x + 2|.

(x + 2)^2 = (0 + 2)^2 = 4
|x + 2| = |0 + 2| = 2
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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 30 Jul 2026, 02:29
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Hi ananya88888888888,

Short answer: there is no third case, because zero is already sitting inside one of the two cases.

To drop the absolute-value bars, the only thing you need to know is the sign of what is inside them. And there are just two options:

- x + 2 is not negative, so |x + 2| = (x + 2)
- x + 2 is negative, so |x + 2| = -(x + 2)

"Not negative" already covers zero. That is the whole answer to your question.

One thing worth separating out: the zero you are asking about is x + 2 = 0, which means x = -2. That is not the same as x = 0, which Bunuel ruled out earlier in the thread - and he is right, because at x = 0 the left side is 4 while the right side is 2. Your zero is a genuine root; that other one is not.

Where the confusion comes from: the post you quoted labels its two branches "positive" and "negative". Zero is neither of those, so it looks like a slot is missing. Write the first branch as x + 2 >= 0 instead of "positive", and the gap closes.

Why it does not matter which branch owns the zero

At x + 2 = 0, both (x + 2) and -(x + 2) equal 0 - the same number. So the zero point gives you the identical equation in either branch, and the identical root. File it under one branch, and just do not file it under both, or you would count that root twice.

Why the wording still matters

In that post, x = -2 shows up in both cases and is kept. That works out, but by luck. If the two ranges were enforced strictly - positive in one, negative in the other - then x = -2 would be out of range in both, and you would lose it. You would be left with (-1)(-3) = 3, which is not even an answer choice.

And -2 really is a solution. Check it: (-2 + 2)^2 = 0, and |-2 + 2| = 0. Both sides are 0.

So the roots are -1, -2, -3, and their product is -6.

Answer: A
ananya88888888888
here shouldnt there be another case which is when it is 0?

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