solvefor x,12x^5=(4x^2)(C)

Lösung

löse nach

x, 12 x^{5} = (4 x^{2}) (C)

x = 0, x = 3 \frac{C}{3}, x = - \frac{3 \frac{C}{3}}{2} + \frac{3 \frac{C}{3} 3}{2} i, x = - \frac{3 \frac{C}{3}}{2} - \frac{3 \frac{C}{3} 3}{2} i

Schritte zur Lösung

12 x^{5} = (4 x^{2}) (C)

Verschiebe

(4 x^{2}) (C)

auf die linke Seite

12 x^{5} - 4 x^{2} C = 0

Faktorisiere

12 x^{5} - 4 x^{2} C : 4 x^{2} (3 x^{3} - C)

4 x^{2} (3 x^{3} - C) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

x = 0 or 3 x^{3} - C = 0

Löse

3 x^{3} - C = 0 : x = 3 \frac{C}{3}, x = - \frac{3 \frac{C}{3}}{2} + \frac{3 \frac{C}{3} 3}{2} i, x = - \frac{3 \frac{C}{3}}{2} - \frac{3 \frac{C}{3} 3}{2} i

Die Lösungen sind

x = 0, x = 3 \frac{C}{3}, x = - \frac{3 \frac{C}{3}}{2} + \frac{3 \frac{C}{3} 3}{2} i, x = - \frac{3 \frac{C}{3}}{2} - \frac{3 \frac{C}{3} 3}{2} i

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

so l v e f or y, y^{4} = x^{3} - 16 x

x^{6} - 2 x^{3} + 1 = 0

(x^{2} - 4) (x^{2} + 1) (x - 3) = 0

3 x^{3} + 5 x^{2} - 4 x - 4 = 0