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this one is tricky and freaky. But easy. Find all its

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this one is tricky and freaky. But easy. Find all its [#permalink]

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19 Jun 2003, 00:08
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this one is tricky and freaky. But easy.
Find all its solutions

(2C)!–|C|=C

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19 Jun 2003, 09:37
Alas, you are wrong again. How did you manage to be admitted by doing so many mistakes? What about 1/2?

For those who still doubt I can post a solution.

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24 Jul 2003, 21:07
How does one know the one got all the possible solutions to such equations? Are there any rules ?

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24 Jul 2003, 22:28
(2C)!=C+|C|

Since C is inside a factorial, C >=0, so open up a modul.

(2C)!=C+C=2C
Let's 2C=A, we have A!=A; thus A=1 or 2

2C=1, thus C=1/2
2C=2, thus C=1

no other roots!

simple, isn't it?

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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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25 Jul 2003, 01:00
How does one know the one got all the possible solutions to such equations? Are there any rules ?

Hi!

Yes. There are rules.

I teach, IMHO, a very good and systematic method to solve ANY type of absolute value problem. Learn this method because it works EVERY TIME.

1) first determine the "critical points" of the equation. The critical points are those point at which a complete expression inside one of the absolute value sign changes from positive to negative or vice-versa.

2) Rewrite the equations in the intervals between +infinity, -infinity, and all of the critical points WITHOUT absolute value signs.
3) Solve for each interval. IF the solution(s) are not compatible with the interval you are in, then that solution is not valid.

Repeat steps 2 and 3 for each interval.

I know it's a little confusing, but let me give you a few examples:

Solve: |x - 1| = 4

Step 1:
The critical point here is x=1 because at this point, the expression x - 1 changes sign.

Steps 2 and 3:
The important intervals are x <= 1 and x > 1. (It is okay to include the endpoint in the intervals).
When x < 1, the expression x - 1 is always negative, so the absval signs will "flip" it. So for this interval, lets rewrite it as:
-(x - 1) = 1 - x = 4
Solving for x, we get x = -3 which is in the interval so it is a valid solution.
When x > 1, the expression x - 1 is always positive, so the absval signs will have no effect on the expression. So for this interval, lets rewrite it as:
x - 1 = 4, and solving, we get x = 5. The solution is in the interval (x > 1) so it is also a valid solution, thus the solution set is x = {-3,5}.

Example 2: (You can do this one in your head only if you are Russian or Chinese (they learn Calculus before puberty), but it's real easy to get confused).

Solve: |x тАУ 2| - |x тАУ 3| = |x тАУ 5|?

Step 1:
Recall, the critical points are -inf, +inf, and any finite numbers that causes any complete expression within an absolute value pair to change signs from negative to positive or vice-versa. For example, as x goes from less than 2 to greater than 2, тАЬx тАУ 2тАЭ тАУ the expression inside the first pair of absolute value signs тАУ goes from negative to positive. Hence, 2 is one of the critical points. By the same logic, we can deduce that the other two critical points are 3 and 5.

Steps 2 and 3:
Since the finite critical point is 2, we will choose our first interval of interest as -inf < x <= 2. In this interval, we now rewrite the equation so that it would evaluate correctly without absolute values signs, i.e., since x - 2 in the interval is always negative, we can rewrite |x тАУ 2| as -(x - 2) or (2 - x). Similarly, bot "x - 3" and "x - 5" will also always be negative if x < 2. We now solve the equation тАЬ(2 тАУ x) тАУ (3 тАУ x) = 5 тАУ xтАЭ subject to the constraint that x lies within the interval -inf < x <= 2. We are able to obtain the solution x = 6. The solution, however, lies outside the interval so it is not considered a valid solution. Hence, we conclude that there are no solutions within this interval (remember, we are solving the equation for x SUCH THAT x is within the interval of interest).

The next interval of interest is between 2 and the next highest finite endpoint, or 2 <= x <= 3. Once again, we rewrite the equation so that we can eliminate the absolute value signs for xтАЩs within the interval and we end up solving for the equation: (x тАУ 2) тАУ (3 тАУ x) = 5 тАУ x. Solving for x, we once again obtain a solution, x = 10/3 = 3.33333тАж, that is outside the interval, so again it is not considered a valid solution.

Similarly, the next interval is 3 <= x <= 5. We solve for the equation: (x тАУ 2) тАУ (x тАУ 3) = 5 тАУ x. This time, the solution x = 4 DOES lie within the interval 3 <= x <= 5, hence it is a valid solution.

Since 5 is our greatest finite critical point, the last interval is 5 <= x < +infinity. We solve for the equation: (x тАУ 2) тАУ (x тАУ 3) = x тАУ 5 obtain the result, x = 6. This solution lies within the interval 5 <= x < +infinity, hence it is also a valid solution.

We conclude that the only two solutions of the original equation are x = 4 and x = 6, so x = {4, 6}.

Example 3: Stolyar's current problem

Solve: (2C)!=C+|C|

Step 1:
There is only one expression inside absolute value signs |C| so the critical point here is clearly C = 0.

Step 2:
The first interval is C <= 0. Rewriting the equation so that it would always be true in this interval, we get (2C)! = C + -C or (2C!) = 0. Since 2C! can never be zero, we conclude that there are no solutions in this interval.

The second interval is C >= 0. Rewriting the equation so that it would always be true in this interval, we get

(2C)! = C + C or (2C!) = 2C.

Now divide both sides by 2C, we get:

(2C - 1)! = 1.

Well, (2C - 1)! will equal 1 only when (2C - 1) = 0 or 1, hence C must equal either 1/2 or 1. Hence, the solution is C = {1/2, 1}.

Note: it is possible to write equations that make this method a little tedious (i've seen a few of Stoylar's equation where there are nested abs val signs such as: ||X^2| - 2X| = ||X + 4| - (|X| - 3)| -- yes, he can be a sadist.)

Two relevant comments: 1) you can still use this method -- it's just a little trickier to determine what the critical points are; and 2) you will NEVER get a problem like this on the GMAT, so don't worry, be happy.

_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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25 Jul 2003, 01:00
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