JeffTargetTestPrep wrote:
viktorija wrote:
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
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!
Bunuel wrote:
SOLUTIONIf \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?A. 5
B. 10
C. 11
D. 20
E. 21
When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\).
1 integer value of \(y\) for this range.
When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\).
Therefore for this range \(-10<(y=2x)<10\).
19 integer values of \(y\) for this range (from -9 to 9, inclusive).
When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).
Hence in this case \(y=|x+5|-|x-5|=x+5-(x-5)=10\).
1 integer value of \(y\) for this range.
Total = 1 + 19 + 1 = 21.
Answer: E.
If anyone interested here is a graph of \(y=|x+5|-|x-5|\):
As you can see y is a continuous function from -10 to 10, inclusive.
Try NEW absolute value
DS question.
Attachment:
WolframAlpha--yx5-x-5_--2014-07-08_0728.png