MATHS F.4......THX!!!

51.)
(x^2 - 2x)^2 - 7(x^2 - 2x) = 8
(x^2 - 2x)^2 - 7(x^2 - 2x) - 8 = 0
[(x^2 - 2x) - 8][(x^2 - 2x) + 1] = 0
x^2 - 2x - 8 = 0 or x^2 - 2x + 1 = 0
(x-4)(x+2) = 0 or (x-1)^2 = 0
x = 4 or -2 or 1
52.)
(x^2 - x)^2 - 8(x^2 - x) + 12 = 0
[(x^2 - x) - 6][(x^2 - x) - 2] = 0
x^2 - x - 6 = 0 or x^2 - x - 2 = 0
(x-3)(x+2) = 0 or (x-2)(x+1) = 0
x = 3 or -2 or 2 or -1
53.)
(x^2 - 2x)^2 + 2x^2 = 4x + 15
(x^2 - 2x)^2 + 2(x^2 - 2x) - 15 = 0
[(x^2 - 2x) + 5][(x^2 - 2x) - 3] = 0
x^2 - 2x + 5 = 0 (rej.) or x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or -1
54.)
√(2x+1) + 4/√(2x+1) = 4
(2x+1) + 4 = 4√(2x+1)
(2x+1) - 4√(2x+1) + 4 = 0
[√(2x+1) -2]^2 = 0
√(2x+1) = 2
2x + 1 = 4
2x = 3
x = 3/2 (or = 1.5)
55.)
√(x-2)+1 = 6/√(x-2)
(x-2) + √(x-2) = 6
(x-2) + √(x-2) - 6 = 0
[√(x-2) + 3][√(x-2) - 2] = 0
√(x-2) = -3 (rej.) or √(x-2) = 2
x-2 = 4
x = 6
57.)
x^3 (x^3 + 7) = 8
x^6 + 7x^3 - 8 = 0
(x^3 + 8)(x^3 - 1) = 0
x^3 = -8 or x^3 = 1
x = -2 or 1
62.)
x^4 - 3x^2 + 2 = 0
(x^2 - 2)(x^2 - 1) = 0
x^2 = 2 or x^2 = 1
x = -√2 or √2 or -1 or 1
64.)
x-5√x+4 = 0
(√x - 4)(√x - 1) = 0
√x = 4 or 1
x = 16 or 1


2010-08-08 15:58:23 補充：
To 002:
x-5√x+4=0
(√x-1)(√x-4)=0
√x=1 or √x=4
x=1 or x=4 or x=-4 (rejected) <<<< x should be 16. but not 4,
Since 4^2 = 16