(2+i)x^2+(3-i)x+1-3i=0

解

(2 + i) x^{2} + (3 - i) x + 1 - 3 i = 0

x = - 1.24898 \dots - \frac{- 3.49796 \dots + 15.21954 \dots}{4} i, x = 0.24898 \dots + \frac{0.50203 \dots + 15.21954 \dots}{4} i

解答ステップ

(2 + i) x^{2} + (3 - i) x + 1 - 3 i = 0

代用

x = a + bi

(2 + i) (a + bi)^{2} + (3 - i) (a + bi) + 1 - 3 i = 0

拡張

(2 + i) (a + bi)^{2} + (3 - i) (a + bi) + 1 - 3 i : (2 a^{2} - 2 b^{2} - 2 ab + 3 a + b + 1) + i (- 3 - a + a^{2} - b^{2} + 3 b + 4 ab)

(2 a^{2} - 2 b^{2} - 2 ab + 3 a + b + 1) + i (- 3 - a + a^{2} - b^{2} + 3 b + 4 ab) = 0

標準的な複素数形式で

0

を書き換える:

0 + 0 i

(2 a^{2} - 2 b^{2} - 2 ab + 3 a + b + 1) + i (- 3 - a + a^{2} - b^{2} + 3 b + 4 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:

[2 a^{2} - 2 b^{2} - 2 ab + 3 a + b + 1 = 0 - 3 - a + a^{2} - b^{2} + 3 b + 4 ab = 0]

[2 a^{2} - 2 b^{2} - 2 ab + 3 a + b + 1 = 0 - 3 - a + a^{2} - b^{2} + 3 b + 4 ab = 0] : (a = - 1.24898 \dots, a = 0.24898 \dots, b = - \frac{- 3.49796 \dots + 15.21954 \dots}{4} b = \frac{0.50203 \dots + 15.21954 \dots}{4})

代用を戻す

x = a + bi

x = - 1.24898 \dots - \frac{- 3.49796 \dots + 15.21954 \dots}{4} i, x = 0.24898 \dots + \frac{0.50203 \dots + 15.21954 \dots}{4} i