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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 17 May 2024, 10:21
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(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

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What is the product of all the solutions of (x + 2)^2 = |x + 2|? 25 May 2024, 06:51
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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
­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


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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
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­
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
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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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apoo151997
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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