3+in(x^2+1)=2in(x)

Lösung

3 + in (x^{2} + 1) = 2 in (x)

x = 1 + i \frac{3 i n}{in}, x = 1 - i \frac{3 i n}{in}; in \neq = 0

Schritte zur Lösung

3 + in (x^{2} + 1) = 2 in (x)

Schreibe

3 + in (x^{2} + 1)

um:

3 + in x^{2} + in

3 + in x^{2} + in = 2 in x

Verschiebe

2 in x

auf die linke Seite

3 + in x^{2} + in - 2 in x = 0

Schreibe in der Standard Form

a x^{2} + b x + c = 0

in x^{2} - 2 in x + 3 + in = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- ( - 2 in ) \pm ( - 2 in ) ^{2} - 4 in ( 3 + in )}{2 in}

Vereinfache

(- 2 in)^{2} - 4 in (3 + in) : 23 i in

x_{1, 2} = \frac{- ( - 2 in ) \pm 2 3 i in}{2 in}; in \neq = 0

Trenne die Lösungen

x_{1} = \frac{- ( - 2 in ) + 2 3 i in}{2 in}, x_{2} = \frac{- ( - 2 in ) - 2 3 i in}{2 in}

x = \frac{- ( - 2 in ) + 2 3 i in}{2 in} : 1 + i \frac{3 i n}{in}

x = \frac{- ( - 2 in ) - 2 3 i in}{2 in} : 1 - i \frac{3 i n}{in}

Die Lösungen für die quadratische Gleichung sind:

x = 1 + i \frac{3 i n}{in}, x = 1 - i \frac{3 i n}{in}; in \neq = 0

(x + 2) (x + 1) - (x - 1) = 6

0 = - \frac{32}{81} (x^{2} - x - 20)

2 x^{2} - 11 x + 3 = - 2

(a x + b)^{2} = c

x = x^{2} + x - 1