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

Lösung

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

x = \frac{k - 1 + k ( k - 3 )}{k + 1}, x = \frac{k - 1 - k ( k - 3 )}{k + 1}; k \neq = - 1

Schritte zur Lösung

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

Schreibe

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

um:

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

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

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

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

Vereinfache

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

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

Trenne die Lösungen

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

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

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

Die Lösungen für die quadratische Gleichung sind:

x = \frac{k - 1 + k ( k - 3 )}{k + 1}, x = \frac{k - 1 - k ( k - 3 )}{k + 1}; k \neq = - 1

3 t^{2} - \frac{1}{2} = 0

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

1 3^{2} = 5^{2} + x^{2}, x

a^{2} + b^{2} - 70 = 1 2^{2} + x^{2}

x^{2} + 0.5 x - 0.11 = 0