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

Lösung

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

x = \frac{2 k - 1 + 8 k - 7}{2 ( k - 2 )}, x = \frac{2 k - 1 - 8 k - 7}{2 ( k - 2 )}; k \neq = 2

Schritte zur Lösung

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

Schreibe

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

um:

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

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

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

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

Vereinfache

(- 2 k + 1)^{2} - 4 (k - 2) (k - 1) : 8 k - 7

x_{1, 2} = \frac{- ( - 2 k + 1 ) \pm 8 k - 7}{2 ( k - 2 )}; k \neq = 2

Trenne die Lösungen

x_{1} = \frac{- ( - 2 k + 1 ) + 8 k - 7}{2 ( k - 2 )}, x_{2} = \frac{- ( - 2 k + 1 ) - 8 k - 7}{2 ( k - 2 )}

x = \frac{- ( - 2 k + 1 ) + 8 k - 7}{2 ( k - 2 )} : \frac{2 k - 1 + 8 k - 7}{2 ( k - 2 )}

x = \frac{- ( - 2 k + 1 ) - 8 k - 7}{2 ( k - 2 )} : \frac{2 k - 1 - 8 k - 7}{2 ( k - 2 )}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{2 k - 1 + 8 k - 7}{2 ( k - 2 )}, x = \frac{2 k - 1 - 8 k - 7}{2 ( k - 2 )}; k \neq = 2

x^{2} - 2 x - 28 = - 6 x + 4 - 6 x + 4

2 x (x - 14) + 11 (3 + x) = - 22 x + 27 + x^{2}

so l v e f or t, y = v t + \frac{1}{2} g t^{2}

2 a^{2} - 2 a - 3 = 0

co m pl e t es q u a re 2 x^{2} - x + 4