請問這數怎解

代數式 4x^2 - 2x - 5 與 2x^2 + 1 的值互為相反數, 即
4x^2 - 2x - 5 = -(2x^2 + 1)
4x^2 - 2x - 5 + 2x^2 + 1 = 0
6x^2 - 2x - 4 = 0
2(3x^2 - x - 2) = 0
2(3x + 2)(x - 1) = 0
x = -2 / 3 或 1.
2. a^2 + ab - b^2 = 0
設b =/= 0 [否則問 a / b是沒有意思的]
(a^2 + ab - b^2) / b^2 = 0 / b^2
a^2 / b^2 + ab / b^2 - 1 = 0
(a / b)^2 + (a / b) - 1 = 0
設x = a / b. 則 x^2 + x - 1 = 0
x = [-1 ± √(1^2 + 4)] / 2
故 a / b = [-1 ± √5] / 2

2. a^2 + ab - b^2
= 0=(a^2*a + a^2*b)-b^2
=(a^3+a^2*b)-b^2
=(a^3*a^2+a^3*b)-b^2
=(a^5+a^3*b)-b^2
=(a^5*a^3+a^5*b)-b^.................than i don't know