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

Lösung

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

Schritte zur Lösung

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

Ersetze

x = u + v i

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

Schreibe um

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

Schreibe

ib

in der Standard komplexen Form um:

0 + bi

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

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]

[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}

Setze in

x = u + v i

ein

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

z^{2} - (2 + i 3) z - 5 + i = 0

x^{2} + 4 i \cdot x - 1 = 0

a \cdot (a - bi) = 4 + 6 i

2 z^{2} - i z - 1 - 3 i = 0

\frac{a}{1 + i} + \frac{b}{1 + 2 i} = 1