nikhil007 wrote:
VeritasPrepKarishma wrote:
Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0
|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)
Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|
You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3
Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out
but I guess in this example its this concept that we need to understand
|x|= x when x is >= 0,
|x|= -x when x < 0
ok, so based on this understanding I will take a fresh shot, please let me know what's wrong
Quote:
a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)
In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct?
and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format?
in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.
Quote:
b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
again, I get it why LHS is that way, however I still don't get it why we don't have -(8-X) as we need to make sure that the result of this bracket is -tive so |8-x| = -(8-x)
Quote:
c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.
Quote:
d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)
(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason?
(4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason?
(8+x) again same reason as above for this?
I also had one doubt in your blog question.
Complication No 3: on this post
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/(-2x^3 + 17x^2 – 30x) > 0
This is how I understand it,
\(x(-2x^2 + 17x - 30) > 0\) (just took out x common) ok
x(2x – 5)(6 – x) > 0(factoring the quadratic) ok
2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common) ---------> I think in this you took out -1 common to make the second bracket = (x-k) format?
First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly.
When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it?
Say |x| = 4, you still haven't got the value of x. You have the value of |x| only. So you need to remove the mod. Now there are rules to remove the mod.
|x|= x (mod removed) when x is >= 0,
|x|= -x (mod removed) when x < 0
So |x| = 4 to remove the mod, I need to know whether x is positive or negative.
If x >= 0, |x| = x so |x| = 4 = x
We get that x is 4
If x < 0, |x| = -x so |x| = 4 = -x
hence x = -4
So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is -4.
Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.
So you split it into ranges:
x > 3
Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x - 3), (x - 3) will remain positive.
So when x > 3, we can remove the mods without any modification:
(x + 4) + (x-3) = 10
x = 9/2
Since 9/2 is greater than 3, this value of x is acceptable.
-4 < x< 3
For these values of x, (x+4) will always be positive but (x-3) will be negative. So |x - 3| = -(x-3)
(x + 4) - (x-3) = 10
You don't have any such value for x
x < -4
For these values of x, (x+4) and (x-3) will be negative. So |x - 3| = -(x-3) and |x+4| = -(x+4)
-(x + 4) - (x-3) = 10
x = -11/2
Since -11/2 is less than -4, this value of x is also acceptable.
I have discussed how to deal with such questions logically here:
http://www.veritasprep.com/blog/2011/01 ... s-part-ii/As for question with factors that are multiplied (discussed in the 3 links given above),
We know how to deal with (x-a)(x-b)(x-c) > 0 type of questions so we try to bring it that form.
2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by -1)-------> how did you arrive at 2(x-0)? i think it should be just \(2x(x-\frac{5}{2})(x-6) <0\)
(x-0) is nothing but x. I put as (x-0) to make it consistent to the (x-a)(x-b).... form to help you remember that you have to take 0 as a transition point too.
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Karishma
Veritas Prep GMAT Instructor
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