i(x)=-2x^2+5x+1

解答

i (x) = - 2 x^{2} + 5 x + 1

x = - 0.18097 \dots - 0.03161 \dots i, x = 2.68097 \dots - 0.46838 \dots i

求解步骤

i (x) = - 2 x^{2} + 5 x + 1

替代

x = a + bi

i (a + bi) = - 2 (a + bi)^{2} + 5 (a + bi) + 1

乘开

- b + ia = (- 2 a^{2} + 2 b^{2} + 5 a + 1) + i (5 b - 4 ab)

复数仅在实部和虚部均相等时才相等改写为方程组:

[- b = - 2 a^{2} + 2 b^{2} + 5 a + 1 a = 5 b - 4 ab]

[- b = - 2 a^{2} + 2 b^{2} + 5 a + 1 a = 5 b - 4 ab] : (a = - 0.18097 \dots, a = 2.68097 \dots, b = - 0.03161 \dots b = - 0.46838 \dots)

x = a + bi

代回

x = - 0.18097 \dots - 0.03161 \dots i, x = 2.68097 \dots - 0.46838 \dots i

CM = C (1 + t i)

(3 + 2 i)^{2} y = (x + 3 i)^{2}

(4 x - 18.6 - 4 y) = (6 y) i + (4 x - 2) j

\frac{2 - 3 i}{2 a + bi} = 3 + 2 i

(x + 1) (x + i) = 0