ahhhhh I see. Thank you for your help.

My pleasure.
Be forewarned:
If we apply your approach, we must still confirm that the resulting solutions are valid in the original equation.

Example:
(x-3)(x+1) = |x-3|
If we apply your approach to (x-3)(x+1) = x-3, we get the following solutions:
x=3, x=0.
However, x=0 is not a valid solution for (x-3)(x+1) = |x-3|:
(0-3)(0+1) = |0-3|
-3 = 3.
Thus, the equation in blue yields only one valid solution for (x-3)(x+1) = |x-3|:
x=3.
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GMATGuruNY

Yes, I see that. I think it will play it safe and go with the approach that everyone has outlined above.

But even if we take X+2 as nagative, then how X=-2 is valid, it will make the solution 0 ? How is -2 a valid value?

Hi kunalkhanna,

To start, were you comfortable 'simplifying' the given equation down into the following equation...?

(X+2)(X+2) = |X+2|

Looking at this, there's NO WAY that X can be positive. Try TESTing any positive number - the two sides will NEVER equal. So there are NO positive answers.

Looking at the answer choices, they're all within a relatively small distance from 0, so the individual values for X are probably all relatively close to 0 as well. Let's find them with a bit of "brute force"...

X = 0
(0+2)(0+2) = |0+2|?
(2)(2) = 2? NO, 4 does not equal 2, so X = 0 is NOT a solution

X = -1
(-1+2)(-1+2) = |-1+2|
(1)(1) = 1? YES, 1 does equal 1, so X = -1 IS a solution

X = -2
(-2+2)(-2+2) = |-2+2|
(0)(0) = 0? YES, 0 does equal 0, so X= -2 IS a solution

Since we're dealing with a Quadratic and an Absolute Value, there are probably going to be MORE than 2 solutions, so we have to keep looking...

X = -3
(-3+2)(-3+2) = |-3+2|
(-1)(-1) = 1? YES, 1 does equal 1, so X= -3 IS a solution

At this point, if we take the product of these 3 answers we have (-1)(-2)(-3) = -6.

Looking at the 5 answer choices, there CAN'T be any other answers. To go from -6 to -12, "+2" would have to be an answer (but we already know that positive numbers DON'T FIT). To go from -6 to +12, (-2) would have to be used twice, which is not a mathematical option. So -6 MUST be the answer.

GMAT assassins aren't born, they're made,
Rich
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the eqn can be simplified as: x^2 + 4x + 4 = |x+2|
this means:

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

so now we have 2 cases to solve for:

x+2=(x+2)^2 --> 1 = (x+2) --> x=-1

then we have

x+2=-(x+2)^2 --> 1 = -(x+2) --> 1 = -x - 2 --> -x = 3 --> x=-3

so my algebra gave me 2 solutions: x= -1 and -3

the solution of -2 was from OBSERVING THE equation rather than the algebra.
Can anyone tell me how i can get -2 without OBSERVING?

regards

Hi Mansoor50,

Your algebraic approach is incomplete. By trying to 'divide out' the Absolute Value, you actually removed the '-2' solution. The inclusion of an Absolute Value almost always means that there will be more than one solution to an equation (and it's almost always a positive one and a negative one); here though, we also have to deal with a squared-term (which ALSO implies more than one solution). If you are going to approach the question algebraically, then you have to account for the fact that (X+2) could be a positive value OR a negative value. In simple terms, you only did half of the work.

GMAT assassins aren't born, they're made,
Rich
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The secret here is realising that we are calculating based on the UNIQUE solutions.

If we don't then we end up with 4 possible solutions
x=-2
x= -1
x=-3
x= -2

When we multiply this together (-2)*(-1)*(-3)*(-2) = 12

This is a punishment answer (E).

Easy to get caught by this trap.
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VeritasKarishma Bunuel I have always had this doubt but haven't come across a good explanation yet. I would really appreciate your help.

Question: While considering a transition point (i.e. x) for absolute values, how do we decide whether that transition point is taken with in the < case or > case?

Example |x+2|, so would it be option 1 or option 2?
Option 1) x <= 2 & x > 2
or
Option 2) x < 2 & x >= 2

I believe it doesn't really matter for this Q since both case would give -2 and we'll need to ignore it one of the them since -2 will not satisfy the range.

But if we look at this Q: https://gmatclub.com/forum/if-y-x-5-x-5 ... l#p1378885 (Quoted below). I believe we need to be careful about the = sign.

Option 1) x≤−5 & −5<x<5 => Gives 21 solutions (-5<x<5 => -10 < y < 10 => y = 19 values)
Option 2) x < 5 & -5<=x<5 => Gives 22 solutions (-5<=x<5 => -10 <= y < 10 => y = 20 values)

I know Option 2 isn't right since if we take -5, it gives |-5+5| - |-5-5| = 0 - |-10| = -10.
Testing values:
x = -6 => |-6+5|−|-6−5| = |-1| - |-11| = 1 - 11 = -10
x = -4 => |-4+5|−|-4−5| = |1| - |-9| = 1 - 9 = -8

From testing those values, it tells us that -5 should belong with < and not > since all < values give -10 while all > values give something else.

Question: Is there any other easy way to identify this on deciding on whether we should include the = sign with the > case or < case?

Thanks!

EducationAisle GMATNinja VeritasKarishma Bunuel I would really appreciate if someone could help with the above query (https://gmatclub.com/forum/what-is-the- ... l#p2353029). Thanks!

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

Region 1:
x>=-2
x^2 + 4x + 7 = x+2 +3
x^2 + 3x + 2 = 0
(x + 1)(x+2) = 0
x = -1 or -2

Region 2:
x<-2
x^2 + 4x + 7 = -x -2 + 3
x^2 + 5x + 6 = 0
(x+2)( x+3) =0
x = -2 or -3

x = {-1,-2,-3}
Product of all solutions of x = (-1)(-2)(-3) = -6


IMO A
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dcummins

x = {-1,-2,-3}
-2 should not be multiplied twice.
(-1)(-2)(-3) = -6
IMO A
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Yes I know.. thats why I spelled it out that we only calculate unique solutions.
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I'm a bit peeved this question didn't specify "distinct" values.... is this common? Are we meant to assume that we take the non-distinct values when asked in the way this question does?

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Beautiful !

Isn't the question unfair in punishing when it doesn't say 'distinct' ?

Hi altairahmad,

In these sorts of Algebraic equations (that ask for the total number of solutions), the word 'distinct' is implied.

For example, how many solutions does the following equation have:

X^2 + 2X + 1 = 0

Does it have one solution or two solutions? When you factor this down, you'll see...

(X+1)(X+1) = 0
X = -1

Even if you write down "-1" twice, there's still just one solution (re: -1); the same logic applies to this prompt (and you're not allowed to 'count' the -2 solution twice).

GMAT assassins aren't born, they're made,
Rich
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