1+t-4t^2+t^3+t^4=0

Lösung

1 + t - 4 t^{2} + t^{3} + t^{4} = 0

t = 1, t = \frac{- 3 + 5}{2}, t = \frac{- 3 - 5}{2}

Dezimale

t = 1, t = - 0.38196 \dots, t = - 2.61803 \dots

Schritte zur Lösung

1 + t - 4 t^{2} + t^{3} + t^{4} = 0

Schreibe in der Standard Form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

t^{4} + t^{3} - 4 t^{2} + t + 1 = 0

Faktorisiere

t^{4} + t^{3} - 4 t^{2} + t + 1 : (t - 1)^{2} (t^{2} + 3 t + 1)

(t - 1)^{2} (t^{2} + 3 t + 1) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

t - 1 = 0 or t^{2} + 3 t + 1 = 0

Löse

t - 1 = 0 : t = 1

Löse

t^{2} + 3 t + 1 = 0 : t = \frac{- 3 + 5}{2}, t = \frac{- 3 - 5}{2}

Die Lösungen sind

t = 1, t = \frac{- 3 + 5}{2}, t = \frac{- 3 - 5}{2}

x^{3} + 6 x^{2} + 11 x + 6 = 0

x^{4} - x^{2} - 12 = 0

so l v e f or a, 2 (a^{3} + b^{3}) = 7

y^{3} + y - 16 = 0

m^{4} - 4 m^{3} + 8 m^{2} - 8 m + 4 = 0