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

解

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

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

解答ステップ

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

代用

x = a + bi

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

拡張

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

複素数は, その実数部と虚数部が等しい場合にのみ等しくなるequation系として書き換える:

[- 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)

代用を戻す

x = a + bi

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