數學問題 : 求x的值

The restriction is different for different cases:

Case 1: If p <= 1, the restriction is x >= 1 or x <= -1
Case 2: If p > 1, the restriction is x >= √p or x <= -√p

since both x^2 - p and x^2 - 1 must be non-negative.

2012-11-23 15:27:14 補充：
And do we need to find solution of x in terms of p?

2012-11-23 15:28:36 補充：
And x should be non-negative since √(x^2 - p) and √(x^2 - 1) are non-negative

2012-11-26 16:54:00 補充：
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圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov12/Crazyeqn1.jpg


2012-11-26 17:40:12 補充：
For p = 4/3:
x = 2/√3 by the formula x = (4-p)/√[8(2-p)]
Sub back into the original equation:
LHS = RHS = 2/√3
Hence the solution x = (4-p)/√[8(2-p)] is valid for 0 <= p <= 4/3, and no solution for p < 0 and 2 > p > 4/3

翻雷滾天 風卷殘雲 : check check x = 4/3

...... oh, my object is different to the question,
I just find the range of x and p.....

(x^2 - p)^(1/2) + 2(x^2 - 1)^(1/2) = x
==> (x^2 - p)^(1/2) = x - 2(x^2 - 1)^(1/2)
==> x^2 - p = x^2 - 4x(x^2 - 1)^(1/2) + 4(x^2 - 1) ............. [square both sides]
==> 4x(x^2 - 1)^(1/2) = p + 4(x^2 - 1)
==> 16(x^2)(x^2 - 1) = p^2 + 8p(x^2 - 1) + 16(x^2 - 1)^2 ... [square both sides again]
Let y = x^2 - 1, therefore, x^2 = y + 1, so
16y(y + 1) = p^2 + 8py + 16y^2
==> 16y - 8py = p^2
==> 16y - 8py + 16 - 8p = p^2 - 8p + 16
==> (16 - 8p)(y + 1) = (p - 4)^2
==> (16 - 8p)x^2 = (p - 4)^2
As x cannot be negative, therefore,
x = |p - 4|/(16 - 8p)^(1/2) ... [provided that (p < 2) as (16 - 8p) must be positive.]

(x^2-p)^(1/2) + 2(x^2-1)^(1/2) = x

Sub x = 1 :

(1-p)^(1/2) + 2(1-1)^(1/2) = 1

(1-p)^(1/2) = 1

1-p = 1

p = 0

Sub p = 0 :

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

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

2(x^2-1)^(1/2) = 0

x^2-1 = 0

x=1

∴x=1