25x^2+60x+37=0a+bi

解

25 x^{2} + 60 x + 37 = 0 a + bi

x = \frac{b - 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}} + - \frac{- 1 + b ^{2} + 1}{50} i, x = - \frac{b + 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}} - - \frac{- 1 + b ^{2} + 1}{50} i, x = \frac{b - 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1} + \frac{b ^{2} + 1 + 1}{5 2} i, x = - \frac{b + 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1} - \frac{b ^{2} + 1 + 1}{5 2} i

解答ステップ

25 x^{2} + 60 x + 37 = 0 \cdot a + bi

代用

x = u + v i

25 (u + v i)^{2} + 60 (u + v i) + 37 = 0 \cdot a + bi

拡張

(25 u^{2} - 25 v^{2} + 60 u + 37) + i (60 v + 50 uv) = ib

標準的な複素数形式で

ib

を書き換える:

0 + bi

(25 u^{2} - 25 v^{2} + 60 u + 37) + i (60 v + 50 uv) = 0 + bi

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:

[25 u^{2} - 25 v^{2} + 60 u + 37 = 0 60 v + 50 uv = b]

[25 u^{2} - 25 v^{2} + 60 u + 37 = 0 60 v + 50 uv = b] : u = \frac{b - 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}}, u = - \frac{b + 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}}, u = \frac{b - 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1}, u = - \frac{b + 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1}, v = - \frac{- 1 + b ^{2} + 1}{50} v = - - \frac{- 1 + b ^{2} + 1}{50} v = \frac{b ^{2} + 1 + 1}{5 2} v = - \frac{b ^{2} + 1 + 1}{5 2}

代用を戻す

x = u + v i

x = \frac{b - 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}} + - \frac{- 1 + b ^{2} + 1}{50} i, x = - \frac{b + 60 - \frac{- 1 + b ^{2} + 1}{50}}{50 - \frac{- 1 + b ^{2} + 1}{50}} - - \frac{- 1 + b ^{2} + 1}{50} i, x = \frac{b - 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1} + \frac{b ^{2} + 1 + 1}{5 2} i, x = - \frac{b + 6 2 b ^{2} + 1 + 1}{5 2 1 + b ^{2} + 1} - \frac{b ^{2} + 1 + 1}{5 2} i