solvefor x,(1-t^2)x^2-2t^2x-(t^2+1)=0

Lösung

löse nach

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

x = \frac{t ^{2} + 1}{1 - t ^{2}}, x = - 1; t \neq = 1, t \neq = - 1

Schritte zur Lösung

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

Schreibe

(1 - t^{2}) x^{2} - 2 t^{2} x - (t^{2} + 1)

um:

x^{2} - x^{2} t^{2} - 2 x t^{2} - t^{2} - 1

x^{2} - x^{2} t^{2} - 2 x t^{2} - t^{2} - 1 = 0

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- ( - 2 t ^{2} ) \pm ( - 2 t ^{2} ) ^{2} - 4 ( 1 - t ^{2} ) ( - t ^{2} - 1 )}{2 ( 1 - t ^{2} )}

(- 2 t^{2})^{2} - 4 (1 - t^{2}) (- t^{2} - 1) = 2

x_{1, 2} = \frac{- ( - 2 t ^{2} ) \pm 2}{2 ( 1 - t ^{2} )}; t \neq = 1, t \neq = - 1

Trenne die Lösungen

x_{1} = \frac{- ( - 2 t ^{2} ) + 2}{2 ( 1 - t ^{2} )}, x_{2} = \frac{- ( - 2 t ^{2} ) - 2}{2 ( 1 - t ^{2} )}

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

x = \frac{- ( - 2 t ^{2} ) - 2}{2 ( 1 - t ^{2} )} : - 1

Die Lösungen für die quadratische Gleichung sind:

x = \frac{t ^{2} + 1}{1 - t ^{2}}, x = - 1; t \neq = 1, t \neq = - 1

co m pl e t es q u a re x^{2} - 3 x - 2 = 0

- 2 x = - x^{2}

(4 x - 1) (x + 2) = 4 x

3 x^{2} + 6 (x - 5) = 8 (2 x - 7)

6000 (1 + x)^{2} = 6242.4