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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
I am also getting E

First eq also gives values of P as +3/2, -3/2.

Edit ........

Ok I got it. Silly me .. :wall
It is A.

Originally posted by remgeo on 04 May 2006, 06:24.
Last edited by remgeo on 04 May 2006, 07:48, edited 1 time in total.
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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
I am getting A.

(1) |P–|P||=3
if P is +ve then |p-p| = 0 = 3 which makes no sense as 0 != 3
so P must be -ve |-p-p|= 3
-2p = 3 -> p = -2/3 sufficient

(2) 2|P|=3

this means p = +3/2 or -3/2

hence A is suff
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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
minu wrote:
I am getting A.

(1) |P–|P||=3
if P is +ve then |p-p| = 0 = 3 which makes no sense as 0 != 3
so P must be -ve |-p-p|= 3
-2p = 3 -> p = -2/3 sufficient


Not that it matters in this case, but still..

| -2p | = 3
ie. 2p = 3
ie. p = 3/2.

Such mistakes can kill you :)
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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
Don't understand why from 1) p can't be +-3/2.

Can somebody explain.
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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
Natalya Khimich wrote:
Don't understand why from 1) p can't be +-3/2.
Can somebody explain.


Well just plug in numbers. If p>0, then the expression will always be 0.
The graph of |P–|P|| is
-2x, for x<0, and
0, for x>= 0
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Re: Find P. (1) |P|P||=3 (2) 2|P|=3 [#permalink]
It is A indeed:
The first equation has only one solution –P-P=3P=-3/2



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