Maths Solving Equation

x^2 + y^2 + 1 = 2(xy + x + 1) ... (1)
y^3 = x + 1 ... (2)
(1) => x^2 + y^2 + 1 = 2xy + 2x + 2
x^2 - 2xy + y^2 = 2x + 1
(x - y)^2 = 2x + 1; 2x + 1 >= 0 or x >= -1/2
x - y = +/-√(2x + 1)
y = x +/- √(2x + 1)
Sub into (2), [x +/- √(2x + 1)]^3 = x + 1
x^3 +/- 3x^2√(2x + 1) + 3x(2x + 1) +/- (2x+1)√(2x + 1) = x + 1
+/-(3x^2 + 2x + 1)√(2x + 1) = 1 - 2x - 6x^2 - x^3
Squaring both sides,
(3x^2 + 2x + 1)^2(2x + 1) = (1 - 2x - 6x^2 - x^3)^2
(9x^4 + 12x^3 + 10x^2 + 4x + 1)(2x + 1) = 1 - 4x - 8x^2 + 22x^3 + 40x^4 + 12x^5 + x^6
x^6 - 6x^5 + 7x^4 - 10x^3 - 26x^2 - 10x = 0
x(x^5 - 6x^4 + 7x^3 - 10x^2 - 26x - 10) = 0
So x = 0 is a solution
Let f(x) = x^5 - 6x^4 + 7x^3 - 10x^2 - 26x - 10
The graph of f(x) indicates there is another root near x = 5
f'(x) = 5x^4 - 24x^3 + 21x^2 - 20x - 26
Using Newton's method,
Xn+1 = Xn - f(Xn) / f'(Xn)
Choosing X1 = 5
X2 = 5 - f(5) / f'(5) = 5.2672
X3 = X2 - f(X2) / f'(X2) = 5.2237
X4 = X3 - f(X3) / f'(X3) = 5.2223
X5 = X4 - f(X4) / f'(X4) = 5.2223
Therefore the other real root for x is 5.2223 (correct to 4 decimal places)

圖片參考：http://img517.imageshack.us/img517/4031/graphr.png

http://img517.imageshack.us/img517/4031/graphr.png