x^2-(1+2i)x+(2+i)=0

解

x^{2} - (1 + 2 i) x + (2 + i) = 0

x = \frac{1}{2} - \frac{- 2 + 11}{2} i, x = \frac{1}{2} + \frac{2 + 11}{2} i

解答ステップ

x^{2} - (1 + 2 i) x + (2 + i) = 0

代用

x = a + bi

(a + bi)^{2} - (1 + 2 i) (a + bi) + 2 + i = 0

拡張

(a + bi)^{2} - (1 + 2 i) (a + bi) + 2 + i : (a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 2 ab)

(a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 2 ab) = 0

標準的な複素数形式で

0

を書き換える:

0 + 0 i

(a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 2 ab) = 0 + 0 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} - a + 2 b + 2 = 0 1 - b - 2 a + 2 ab = 0]

[a^{2} - b^{2} - a + 2 b + 2 = 0 1 - b - 2 a + 2 ab = 0] : (a = \frac{1}{2}, a = \frac{1}{2}, b = - \frac{- 2 + 11}{2} b = \frac{2 + 11}{2})

代用を戻す

x = a + bi

x = \frac{1}{2} - \frac{- 2 + 11}{2} i, x = \frac{1}{2} + \frac{2 + 11}{2} i