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There is flaw in this solution {Forgive me if I am wrong}
when you are taking modulus negative then x<-2, so after getting the solution as -3 and -2 we have to eliminate the -2 from here.

{The reason for elimination is consider the equation (x-2) ^2 =0} {here if asked the product of the roots then we will say 4 and not just 2, so we have to count the root 2 twice.}

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viktorija
What is the product of all the solutions of x^2+4x+7=|x+2|+3?

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

Let’s first solve for when x + 2 is positive:

x^2 + 4x + 7 = x + 2 + 3

x^2 + 4x + 7 = x + 5

x^2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = -2 or x = -1

Next we can solve for when x + 2 is negative:

x^2 + 4x + 7 = -(x + 2) + 3

x^2 + 4x + 7 = -x + 1

x^2 + 5x + 6 = 0

(x + 3)(x + 2) = 0

x = -3 or x = -2

We see that the product of all the solutions is (-2)(-1)(-3) = -6.

Answer: A

The equation x^2 + 4x + 7 = |x + 2| + 3 has three roots: -3, -2, and -1.

When we open the modulus and assume x ≥ -2, we get x = -2 and x = -1.

When we assume x < -2, we get x = -3. Note that x = -2 must be discarded in the second case, since it does not fall within the range x < -2.

That said, (x - 2)^2 = 0 has only one root, x = 2.
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Thank you Bunuel for the Reply:
I got the GMAT Q Right though,

I was just confused with the posted Solution that is all,
though out of curiosity,
what about the products of the root of this Equation then (x - 2)^2 = 0, isnt it 4 ?
as product of the roots is c/a which will give 4 {which means the roots are 2 and 2 ?}
Bunuel
Nsp10
There is flaw in this solution {Forgive me if I am wrong}
when you are taking modulus negative then x<-2, so after getting the solution as -3 and -2 we have to eliminate the -2 from here.

{The reason for elimination is consider the equation (x-2) ^2 =0} {here if asked the product of the roots then we will say 4 and not just 2, so we have to count the root 2 twice.}


When we open the modulus and assume x ≥ -2, we get x = -2 and x = -1.

When we assume x < -2, we get x = -3. Note that x = -2 must be discarded in the second case, since it does not fall within the range x < -2.

That said, (x - 2)^2 = 0 has only one root, x = 2.
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Nsp10
Thank you Bunuel for the Reply:
I got the GMAT Q Right though,

I was just confused with the posted Solution that is all,
though out of curiosity,
what about the products of the root of this Equation then (x - 2)^2 = 0, isnt it 4 ?
as product of the roots is c/a which will give 4 {which means the roots are 2 and 2 ?}
Bunuel
Nsp10
There is flaw in this solution {Forgive me if I am wrong}
when you are taking modulus negative then x<-2, so after getting the solution as -3 and -2 we have to eliminate the -2 from here.

{The reason for elimination is consider the equation (x-2) ^2 =0} {here if asked the product of the roots then we will say 4 and not just 2, so we have to count the root 2 twice.}


When we open the modulus and assume x ≥ -2, we get x = -2 and x = -1.

When we assume x < -2, we get x = -3. Note that x = -2 must be discarded in the second case, since it does not fall within the range x < -2.

That said, (x - 2)^2 = 0 has only one root, x = 2.

For (x - 2)^2 = 0:

It has one root: x = 2 (called a repeated root, or double root).

If you expand it to x^2 - 4x + 4 = 0 and apply Vieta’s formulas, you get:


Product of roots (c/a) = 4

Sum of roots (-b/a) = 4

This is because Vieta’s formulas count repeated roots separately.

However, the default assumption, unless stated otherwise, is to consider only unique roots when asked for the sum or product of the roots. So unless the question explicitly mentions “counting multiplicities” or similar wording, it refers to distinct roots only.
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Why the solutions are not being tested for validity? I recall on the TTP course, it is said that the solutions involving absolute values should always be tested for validity. Please guide.
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viktorija
What is the product of all the solutions of x^2+4x+7=|x+2|+3?

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

Let’s first solve for when x + 2 is positive:

x^2 + 4x + 7 = x + 2 + 3

x^2 + 4x + 7 = x + 5

x^2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = -2 or x = -1

Next we can solve for when x + 2 is negative:

x^2 + 4x + 7 = -(x + 2) + 3

x^2 + 4x + 7 = -x + 1

x^2 + 5x + 6 = 0

(x + 3)(x + 2) = 0

x = -3 or x = -2

We see that the product of all the solutions is (-2)(-1)(-3) = -6.

Answer: A
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