solvefor t,a^2b^2t^2-a^2t=1-b^2t

Lösung

löse nach

t, a^{2} b^{2} t^{2} - a^{2} t = 1 - b^{2} t

t = \frac{1}{b ^{2}}, t = - \frac{1}{a ^{2}}; a^{2} b^{2} \neq = 0

Schritte zur Lösung

a^{2} b^{2} t^{2} - a^{2} t = 1 - b^{2} t

Verschiebe

b^{2} t

auf die linke Seite

a^{2} b^{2} t^{2} - a^{2} t + b^{2} t = 1

Verschiebe

1

auf die linke Seite

a^{2} b^{2} t^{2} - a^{2} t + b^{2} t - 1 = 0

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

t_{1, 2} = \frac{- ( - a ^{2} + b ^{2} ) \pm ( - a ^{2} + b ^{2} ) ^{2} - 4 a ^{2} b ^{2} ( - 1 )}{2 a ^{2} b ^{2}}

Vereinfache

(- a^{2} + b^{2})^{2} - 4 a^{2} b^{2} (- 1) : a^{2} + b^{2}

t_{1, 2} = \frac{- ( - a ^{2} + b ^{2} ) \pm ( a ^{2} + b ^{2} )}{2 a ^{2} b ^{2}}; a^{2} b^{2} \neq = 0

Trenne die Lösungen

t_{1} = \frac{- ( - a ^{2} + b ^{2} ) + a ^{2} + b ^{2}}{2 a ^{2} b ^{2}}, t_{2} = \frac{- ( - a ^{2} + b ^{2} ) - ( a ^{2} + b ^{2} )}{2 a ^{2} b ^{2}}

t = \frac{- ( - a ^{2} + b ^{2} ) + a ^{2} + b ^{2}}{2 a ^{2} b ^{2}} : \frac{1}{b ^{2}}

t = \frac{- ( - a ^{2} + b ^{2} ) - ( a ^{2} + b ^{2} )}{2 a ^{2} b ^{2}} : - \frac{1}{a ^{2}}

Die Lösungen für die quadratische Gleichung sind:

t = \frac{1}{b ^{2}}, t = - \frac{1}{a ^{2}}; a^{2} b^{2} \neq = 0

43 \cdot x^{2} + 5 x - 23 = 0

d^{2} = 4 x^{2} - 12 x + 9

0.2 x^{2} + 1.2 x - 32 = 0

so l v e f or x, y^{2} e^{1} + x^{2} = e (x^{2} - 1) - 1 e (x^{2} + 1)

9 x^{2} - 25 = d