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what is product of solutions

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what is product of solutions

by clock60 » Fri Jan 14, 2011 10:56 am
hi all
need help solving below problem
what is the product of all solutions of x^2+4x+7=|x+2|+3
a)-6
b)-2
c)2
d)6
e)12
oa is -6


what is out of my understanding is that, then x+2>0 x>-2 we have x=-2,-1, but only possible solution x=-1
then x+2<0, x<-2 possible x are -2.-3 and x=-3 the only solution
so product=(-1)*(-3)=3 and i have no this answer, and i cant spot my mistake
need help
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by aleph777 » Fri Jan 14, 2011 11:16 am
Where'd you find this problem?

You should approach it as a quadratic, but since there's an absolute value, you need to solve the whole formula twice: once for the absolute value as positive, and one for it as negative.

Let's try.

GIVEN: x^2+4x+7=|x+2|+3

1. SOLVE FOR POSITIVE VALUE OF |x+2|:

x^2+4x+7=x+2+3
x^2+4x+7=x+5
x^2+3x+2=0
And then factor the quadratic...
(x+2)(x+1)=0
Therefore:
x=-2 or x=-1

2. SOLVE FOR NEGATIVE VALUE OF |x+2|:

x^2+4x+7=-x-2+3
x^2+4x+7=-x+1
x^2+5x+6=0
And then factor the quadratic...
(x+3)(x+2)=0
Therefore:
x=-3 or x=-2

I then multiplied "all solutions," as the question asked:

-3 * -2 * -1 * -2 = 12

I answered E, but OA says answer a, which is -6. This means they only counted -2 once, since it appears in both solutions.

I asked myself whether or not to count it twice, but I did because I figured the question would ask "what is the product of all UNIQUE solutions" if it only wanted us to count each number one time...
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by clock60 » Fri Jan 14, 2011 11:25 am
hi aleph777, unfortunately the source is reliable and the answer is a it is not a typo
but for your solution
you said that SOLVE FOR POSITIVE VALUE OF |x+2|:
and got -2 insert this in the |x+2|=|-2+2|=0. 0 is not greater then 0
i mean that
|x|=x if x>0 and -x is x<0 it can be greater or equal
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by aleph777 » Fri Jan 14, 2011 12:01 pm
Clock60,

I'm not sure what your question is, though. It's not an inequality problem. If you plug in -2 for the whole equation you get:

-2^2+4(-2)+7=|-2+2|+3
4+(-8) +7=0+3
3=3
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by clock60 » Fri Jan 14, 2011 12:17 pm
yes friend you are right, i was so involved in inequality that start to treat all with restrictions in this case the problem looks solvable (-1)*(-2)(-3)=-6
you don`t need to multiply (-2) twice as all possible values -1,-2,-3
thanks
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by aleph777 » Fri Jan 14, 2011 12:19 pm
ha! sounds like you could use a little 10 minute snack break!
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by top_business_2011 » Fri Jul 08, 2011 11:19 pm
clock60 wrote:hi all
need help solving below problem
what is the product of all solutions of x^2+4x+7=|x+2|+3
a)-6
b)-2
c)2
d)6
e)12
oa is -6


what is out of my understanding is that, then x+2>0 x>-2 we have x=-2,-1, but only possible solution x=-1
then x+2<0, x<-2 possible x are -2.-3 and x=-3 the only solution
so product=(-1)*(-3)=3 and i have no this answer, and i cant spot my mistake
need help
Hi Guys, I wanted to chime in to bridge a concept gap I sensed.

@ clock60, your initial take on the question was correct except that you reckoned without one additional condition.

While working on problems involving absolute values, it is essential to bear the following conditions in mind:
|x+2| = x+2, if x>-2
-x-2, if x<-2
0, if x=-2
Likewise, the question shall be addressed along all these possible conditions.
x^2+4x+7=|x+2|+3
Case 1: If x>-2,
x^2 + 4x + 7 = x + 2 + 3
Solving this shall give you x = -2 or x = -1
However, the condition we are working with necessitates that an 'x' value be greater than -2; thus the only valid answer is -1.
Case 2: If x<-2,
x^2 + 4x + 7 = -x -2 + 3
Solving this shall give you x = -2 or x = -3
With similar reasoning, x = -2 is not a valid answer as the condition requires the value be less than -2. So the only answer here is -3
Case 3: If x = 2,
x^2 + 4x + 7 = 0 + 3
x^2 + 4x + 7 -3 =0
Solving this gives you x = -2
So, here the value of x is in line with the condition we're working with. As such, a valid answer in this case is -2

Finally, the product of the three valid solutions is -6 [ -1 * -3 * -2]

The concept presented above addresses, maybe, one of the commonest errors test takers commit in the GMAT.

Hope that helps.
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by goalevan » Sat Jul 09, 2011 1:47 pm
When I have an absolute value expression isolated on one side with an expression in the same variable on the other side, I find that making a quick graph can greatly simplify things. After drawing up the chart, I can quickly pick out the solutions to the original equation: -3, -2, and -1.
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