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

解

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

x = 0.20216 \dots - 0.67528 \dots i, x = - 7.20216 \dots + 0.67528 \dots i

解答ステップ

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

代用

x = a + bi

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

拡張

(a + bi)^{2} + 5 (a + bi) + 2 (a + bi) : (a^{2} - b^{2} + 7 a) + i (7 b + 2 ab)

(a^{2} - b^{2} + 7 a) + i (7 b + 2 ab) = 1 - 5 i

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:

[a^{2} - b^{2} + 7 a = 1 7 b + 2 ab = - 5]

[a^{2} - b^{2} + 7 a = 1 7 b + 2 ab = - 5] : (a = 0.20216 \dots, a = - 7.20216 \dots, b = - 0.67528 \dots b = 0.67528 \dots)

代用を戻す

x = a + bi

x = 0.20216 \dots - 0.67528 \dots i, x = - 7.20216 \dots + 0.67528 \dots i