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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

Let, \(|x + 2| = a\)

i.e. equation \(|x + 2|^2 - 5|x + 2| = -6\) becomes \(a^2 - 5a = -6\)

i.e. \(a^2 - 5a +6 = 0\)

i.e. \(a = 3, 2\)

If \(a = 3 = |x + 2|\) then \(x = 1 or -5\)

If \(a = 2 = |x + 2|\) then \(x = 0 or -4\)

Product of all possible values of \(x = (1)*(-5)*(0)*(-4) = 0\)

Answer: Option D
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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

Let, \(|x + 2| = a\)

i.e. equation \(|x + 2|^2 - 5|x + 2| = -6\) becomes \(a^2 - 5a = -6\)

i.e. \(a^2 - 5a +6 = 0\)

i.e. \(a = 3, 2\)

If \(a = 3 = |x + 2|\) then \(x = 1 or -5\)

If \(a = 2 = |x + 2|\) then \(x = 0 or -4\)

Product of all possible values of \(x = (1)*(-5)*(0)*(-4) = 0\)

Answer: Option D

Hey Insight,

Hope you're well.

In a question like this, why can't I multiply out the brackets to make a quadratic equation then solve for x? For example, once I multiplied and simplified the above, I deduced: x^2 + 4x - 5 = 0, where x = -5 or 1. Therefore, the answer would be -5.

May you point out the flaw in my reasoning please?

Thanks in advance.
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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

Let, \(|x + 2| = a\)

i.e. equation \(|x + 2|^2 - 5|x + 2| = -6\) becomes \(a^2 - 5a = -6\)

i.e. \(a^2 - 5a +6 = 0\)

i.e. \(a = 3, 2\)

If \(a = 3 = |x + 2|\) then \(x = 1 or -5\)

If \(a = 2 = |x + 2|\) then \(x = 0 or -4\)

Product of all possible values of \(x = (1)*(-5)*(0)*(-4) = 0\)

Answer: Option D

Hey Insight,

Hope you're well.

In a question like this, why can't I multiply out the brackets to make a quadratic equation then solve for x? For example, once I multiplied and simplified the above, I deduced: x^2 + 4x - 5 = 0, where x = -5 or 1. Therefore, the answer would be -5.

May you point out the flaw in my reasoning please?

Thanks in advance.

It's not bracket (parenthesis) it's modulus sign which considers absolute value i.e. the part inside may be positive as well as negative. You have skipped the possibility of the x+2 to be negative when you treat it like a bracket.
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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

We always say that you need to keep an eye on the options from the beginning. Here, I see an equation with absolute values and we are looking for its solutions.
The product of the solutions look like easy numbers (I think -4, 5 etc).
I notice 0 and the first thing I do here is check for 0. Does x = 0 work? If yes, the product will be 0 no matter what the other values of x are.
x = 0 works!
We are done.

Answer (D)
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What is the product of all possible solutions of the equation |x+2|2−5|x+2|=−6|x+2|2−5|x+2|=−6?

A. -20
B. -5
C. -4
D. 0
E. 20

Let |x+2|= Y
y^2-5y+6=0

solve for y = 2,3

since |x+2|= Y
we get values, 0,1,2,3
product would be 0 hence D
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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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Bunuel
What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

We always say that you need to keep an eye on the options from the beginning. Here, I see an equation with absolute values and we are looking for its solutions.
The product of the solutions look like easy numbers (I think -4, 5 etc).
I notice 0 and the first thing I do here is check for 0. Does x = 0 work? If yes, the product will be 0 no matter what the other values of x are.
x = 0 works!
We are done.

Answer (D)

So it looks like in this instance if you plug in 0, it works. But doing that presupposes that x = 0. Isn't the question asking what is the product of xy?

Actually...so x = 0 ... xy = 0.
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Can somebody please show me how to insert the values of 3 and 2, to get the the four values of x?

I mean this step:


If a=3=|x+2|a=3=|x+2| then x=1or−5x=1or−5

If a=2=|x+2|a=2=|x+2| then x=0or−4
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Champer21
Can somebody please show me how to insert the values of 3 and 2, to get the the four values of x?

I mean this step:


If a=3=|x+2|a=3=|x+2| then x=1or−5x=1or−5

If a=2=|x+2|a=2=|x+2| then x=0or−4

We denoted |x + 2| as a, and then got that a = 3 or a = 2. Hence, |x + 2| = 3 or |x + 2| = 2.

If |x + 2| = 3, then either x + 2 = 3 or x + 2 = -3. Thus, x = 1 or x = -5.
If |x + 2| = 2, then either x + 2 = 2 or x + 2 = -2. Thus, x = 0 or x = -4.

Hope it's clear.
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Bunuel
Champer21
Can somebody please show me how to insert the values of 3 and 2, to get the the four values of x?

I mean this step:


If a=3=|x+2|a=3=|x+2| then x=1or−5x=1or−5

If a=2=|x+2|a=2=|x+2| then x=0or−4

We denoted |x + 2| as a, and then got that a = 3 or a = 2. Hence, |x + 2| = 3 or |x + 2| = 2.

If |x + 2| = 3, then either x + 2 = 3 or x + 2 = -3. Thus, x = 1 or x = -5.
If |x + 2| = 2, then either x + 2 = 2 or x + 2 = -2. Thus, x = 0 or x = -4.

Hope it's clear.

Thank you very much Bunuel. It is not normal how fast you reply.
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What is the product of all possible solutions of the equation \(|x + 2|^2 - 5|x + 2| = -6\)?

A. -20
B. -5
C. -4
D. 0
E. 20

Since its a polynomial of degree two, we will get two answers - unless we have zero. However, it is absolute value that gets two values so any variable inside the absolute would have four values - unless we have zero.

Let's say y = |x+2|
Therefore,
\(y^2 - 5y = -6\)
gives y = 3 and 2

|x+2| = 3 gives x = 1 and -5
|x+2| = 2 gives x = 0

Now, this gives product as zero.

Answer D.
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Is it wrong to treat the absolute value like parentheses?
I considered it as parentheses and got -
x^2 + 4 + 4x -5x - 10 = -6
x^2 -x = 0
x(x-1) = 0

Therefore, x = 1 and x = 0.
On multiplication of the roots 1 and 0 , you get 0.
ans (D).

Is this approach wrong? If yes, why?
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Anu2021
Is it wrong to treat the absolute value like parentheses?
I considered it as parentheses and got -
x^2 + 4 + 4x -5x - 10 = -6
x^2 -x = 0
x(x-1) = 0

Therefore, x = 1 and x = 0.
On multiplication of the roots 1 and 0 , you get 0.
ans (D).

Is this approach wrong? If yes, why?

Absolute value and parentheses are not the same.

Absolute Value



For more check Ultimate GMAT Quantitative Megathread

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