(0.0012+x)(2x)^2=2.8x10^{-7}

Lösung

(0.0012 + x) (2 x)^{2} = 2.8 x 1 0^{- 7}

x = 0, x = \frac{- 0.0012 + 0.00000172}{2}, x = \frac{- 0.0012 - 0.00000172}{2}

Dezimale

x = 0, x = 0.00005 \dots, x = - 0.00125 \dots

Schritte zur Lösung

(0.0012 + x) (2 x)^{2} = 2.8 x \cdot 1 0^{- 7}

Verschiebe

2.8 x 1 0^{- 7}

auf die linke Seite

(0.0012 + x) (2 x)^{2} - 2.8 x \cdot 1 0^{- 7} = 0

Faktorisiere

(0.0012 + x) (2 x)^{2} - 2.8 x \cdot 1 0^{- 7} : x (4 x^{2} + 0.0048 x - \frac{2.8}{10000000})

x (4 x^{2} + 0.0048 x - \frac{2.8}{10000000}) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

x = 0 or 4 x^{2} + 0.0048 x - \frac{2.8}{10000000} = 0

Löse

4 x^{2} + 0.0048 x - \frac{2.8}{10000000} = 0 : x = \frac{- 0.0012 + 0.00000172}{2}, x = \frac{- 0.0012 - 0.00000172}{2}

Die Lösungen sind

x = 0, x = \frac{- 0.0012 + 0.00000172}{2}, x = \frac{- 0.0012 - 0.00000172}{2}

(x - 1) (2 x^{2} - 5 x + 2) = 0

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

r^{5} + 5 r^{4} - 1 = 0

(a - 2) (a + 2) - (a - 2)^{2} a = 1

(x^{2} - 3 x + 3) (x^{2} - 2 x + 3) = 2 x^{2}