(k-1)x^2+2x+(1-k)=0

Lösung

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

x = \frac{- 1 + - k ^{3} + 3 k ^{2} - 3 k + 2}{k - 1}, x = - \frac{- k ^{3} + 3 k ^{2} - 3 k + 2 + 1}{k - 1}; k \neq = 1

Schritte zur Lösung

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

Schreibe

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

um:

k x^{2} - x^{2} + 2 x + 1 - k

k x^{2} - x^{2} + 2 x + 1 - k = 0

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

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

Vereinfache

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

x_{1, 2} = \frac{- 2 \pm 2 - k ^{3} + 3 k ^{2} - 3 k + 2}{2 ( k - 1 )}; k \neq = 1

Trenne die Lösungen

x_{1} = \frac{- 2 + 2 - k ^{3} + 3 k ^{2} - 3 k + 2}{2 ( k - 1 )}, x_{2} = \frac{- 2 - 2 - k ^{3} + 3 k ^{2} - 3 k + 2}{2 ( k - 1 )}

x = \frac{- 2 + 2 - k ^{3} + 3 k ^{2} - 3 k + 2}{2 ( k - 1 )} : \frac{- 1 + - k ^{3} + 3 k ^{2} - 3 k + 2}{k - 1}

x = \frac{- 2 - 2 - k ^{3} + 3 k ^{2} - 3 k + 2}{2 ( k - 1 )} : - \frac{- k ^{3} + 3 k ^{2} - 3 k + 2 + 1}{k - 1}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- 1 + - k ^{3} + 3 k ^{2} - 3 k + 2}{k - 1}, x = - \frac{- k ^{3} + 3 k ^{2} - 3 k + 2 + 1}{k - 1}; k \neq = 1

co m pl e t es q u a re 2 x^{2} - 6 x + 1 = 0

x^{2} + 0.5 x - 0.5 = 0

4 x^{2} - 16 x - 33 = 0

9 x^{2} - 24 x + 20 = 0

x^{2} = 2^{2} + 3^{2}