VeritasPrepKarishma wrote:
In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
Hi Karishma
I went through your post on the blog, but to be frank found this post of your more helpfull
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?
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 \(2x(x-\frac{5}{2})(x-6) <0\)