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# Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0

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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 [#permalink]

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24 Apr 2006, 11:43
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Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
Manager
Joined: 20 Nov 2004
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24 Apr 2006, 12:33
From (1):

|x + 1| = 2|x - 1|
|x+1|^2 = (2|x - 1|)^2
--> this is valid cause both are positive. No solutions will be lost.

x^2 + 2 x + 1 = 4 (x^2 - 2 x + 1)

3x^2 - 10x + 3 = 0

x = 3 OR x = 1/3

Not sufficient.

From (2)
|x - 3| ≠ 0
x ≠ 3

Not sufficient.

(1) and (2) taken together:

Sufficient. Only x = 1/3 survives.

So it's C.
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28 Apr 2006, 23:34
ccax wrote:
From (1):

|x + 1| = 2|x - 1|
|x+1|^2 = (2|x - 1|)^2
--> this is valid cause both are positive. No solutions will be lost.

x^2 + 2 x + 1 = 4 (x^2 - 2 x + 1)

3x^2 - 10x + 3 = 0

x = 3 OR x = 1/3

Not sufficient.

From (2)
|x - 3| ≠ 0
x ≠ 3

Not sufficient.

(1) and (2) taken together:

Sufficient. Only x = 1/3 survives.

So it's C.

_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

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29 Apr 2006, 19:07
ccax wrote:
From (1):

|x + 1| = 2|x - 1|
|x+1|^2 = (2|x - 1|)^2
--> this is valid cause both are positive. No solutions will be lost.

x^2 + 2 x + 1 = 4 (x^2 - 2 x + 1)

3x^2 - 10x + 3 = 0

x = 3 OR x = 1/3

Not sufficient.

From (2)
|x - 3| ≠ 0
x ≠ 3

Not sufficient.

(1) and (2) taken together:

Sufficient. Only x = 1/3 survives.

So it's C.

agreed, nice solution. do we need to know the quadratic formula to solve complex quadratic equations?
29 Apr 2006, 19:07
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