i(x)=-x^2+60x

解答

i (x) = - x^{2} + 60 x

x = 0, x = 60 - i

求解步骤

i (x) = - x^{2} + 60 x

替代

x = a + bi

i (a + bi) = - (a + bi)^{2} + 60 (a + bi)

乘开

- b + ia = (- a^{2} + b^{2} + 60 a) + i (60 b - 2 ab)

复数仅在实部和虚部均相等时才相等改写为方程组:

[- b = - a^{2} + b^{2} + 60 a a = 60 b - 2 ab]

[- b = - a^{2} + b^{2} + 60 a a = 60 b - 2 ab] : (a = 0, a = 60, b = 0 b = - 1)

x = a + bi

代回

x = 0, x = 60 - i

so l v e f or m, (m - n)^{2} = - 2 i

k_{1} - 3 - 27 i k_{2} = 0

so l v e f or c, i = \frac{d}{( 1 - c )}

(a + i)^{2} = 8 + bi

\frac{3 + ai}{5 + bi} = 3 + 5 i