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# PS: simple equation 3

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SVP
Joined: 03 Feb 2003
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02 Jul 2004, 01:11
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||Z|-2|=Z-1

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Intern
Joined: 26 Jan 2003
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Re: PS: simple equation 3 [#permalink]

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02 Jul 2004, 05:09
stolyar wrote:
||Z|-2|=Z-1

this one is a bit harder.

Z >= 1
|Z| - 2 = Z - 1
|Z| = Z + 1
Z = -1/2, but it does't correspond to our range (Z>=1)

no solution?
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Intern
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02 Jul 2004, 08:12
||z|-2| = z-1

Squaring both sides

|z|^2 - 4|z| + 4 = z^2 - 2z + 1

Since |z|^2 = z^2 we get

-4|z| + 4 = -2z + 1
4|z| = 2z + 3

Again squaring both sides

16z^2 = 4z^2 + 12z + 9

ie 4z^2 - 4z - 3 = 0

Solving for Z we get z = -1/2 or z = 3/2

But when choices are given, checking by substitution will be easier than this long process.

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SVP
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02 Jul 2004, 10:33
sathya76 wrote:
||z|-2| = z-1

Squaring both sides

|z|^2 - 4|z| + 4 = z^2 - 2z + 1

Since |z|^2 = z^2 we get

-4|z| + 4 = -2z + 1
4|z| = 2z + 3

Again squaring both sides

16z^2 = 4z^2 + 12z + 9

ie 4z^2 - 4z - 3 = 0

Solving for Z we get z = -1/2 or z = 3/2

But when choices are given, checking by substitution will be easier than this long process.

-1/2 is clearly out. When you square an equation, you can get many extraneous roots. Therefore, each root should be plugged to make sure that it does make sense.

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Senior Manager
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02 Jul 2004, 15:53
boksana wrote:
3/2

agree....with 3/2

knowing z>=1 , so |z|=z.

now equation reduces to

|z-2| = z-1

squarring and reducing , we get z=3/2 as the only solution.

- ash
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Senior Manager
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07 Jul 2004, 06:39
I think the answer should be -1/2 and 3/2.

(Just FYI, Plug these two in original equation and you will get the perfect answer)

The way I did it.

From ||Z|-2|=Z-1
a) |Z|-2 = Z-1 or b) |Z|-2 = -Z + 1

From a)
i) Z-2 = Z-1 or ii) -Z-2 = Z-1
i) not possible
ii) gives z=-1/2

From b)
i) Z-2=-Z+1 or ii) -Z-2= -Z+1
i) gives Z=3/2
ii) not possible.

Any suggestions???

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07 Jul 2004, 06:39
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