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

Lösung

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

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

Schritte zur Lösung

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

Ersetze

x = u + v i

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

Schreibe um

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

Schreibe

i (b + ab)

in der Standard komplexen Form um:

0 + (b + ab) i

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

Komplexe Zahlen können nur gleich sein wenn ihre realen und imaginären Teile gleich sind.Schreibe als Gleichungssystem um

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

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

Setze in

x = u + v i

ein

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

∣ Z + 1 ∣ = (Z + 2) (1 + i)

1 + 2 i = z + \frac{i}{z}

(x - y) + 3 i = 2 + y i

2 x + 2 y = 2 x + 2 y i + 10, x, y

z^{2} + (2 + 5 i) \cdot z + (- 6 + 6 i) = 0