4b^4=2b

Lösung

4 b^{4} = 2 b

b = 0, b = \frac{1}{3 2}, b = - \frac{2 ^{\frac{2}{3}}}{4} + i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}, b = - \frac{2 ^{\frac{2}{3}}}{4} - i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}

Schritte zur Lösung

4 b^{4} = 2 b

Verschiebe

2 b

auf die linke Seite

4 b^{4} - 2 b = 0

Faktorisiere

4 b^{4} - 2 b : 2 b (32 b - 1) (2^{\frac{2}{3}} b^{2} + 32 b + 1)

2 b (32 b - 1) (2^{\frac{2}{3}} b^{2} + 32 b + 1) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

b = 0 or 32 b - 1 = 0 or 2^{\frac{2}{3}} b^{2} + 32 b + 1 = 0

Löse

32 b - 1 = 0 : b = \frac{1}{3 2}

Löse

2^{\frac{2}{3}} b^{2} + 32 b + 1 = 0 : b = - \frac{2 ^{\frac{2}{3}}}{4} + i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}, b = - \frac{2 ^{\frac{2}{3}}}{4} - i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}

Die Lösungen sind

b = 0, b = \frac{1}{3 2}, b = - \frac{2 ^{\frac{2}{3}}}{4} + i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}, b = - \frac{2 ^{\frac{2}{3}}}{4} - i \frac{3 2 3 2 ^{\frac{2}{3}}}{4}

4 x^{3} - 6 x^{2} - 18 x + 27 = 0

x^{3} + 9 x - 6 = 0

- x^{4} + 2 x^{3} + x^{2} - 4 x + 2 = 0

x^{4} - x^{3} - 8 x^{2} - 4 x - 48 = 0

(x^{2} + 1) + (x^{2} + 1) (x^{4} + 1) = 0