(k-1)x^2+2x+(2k-3)=0

Lösung

(k - 1) x^{2} + 2 x + (2 k - 3) = 0

x = \frac{- 1 + - 2 k ^{2} + 5 k - 2}{k - 1}, x = - \frac{- 2 k ^{2} + 5 k - 2 + 1}{k - 1}; k \neq = 1

Schritte zur Lösung

(k - 1) x^{2} + 2 x + (2 k - 3) = 0

Schreibe

(k - 1) x^{2} + 2 x + (2 k - 3)

um:

k x^{2} - x^{2} + 2 x + 2 k - 3

k x^{2} - x^{2} + 2 x + 2 k - 3 = 0

Schreibe in der Standard Form

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

(k - 1) x^{2} + 2 x + 2 k - 3 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( k - 1 ) ( 2 k - 3 )}{2 ( k - 1 )}

Vereinfache

2^{2} - 4 (k - 1) (2 k - 3) : 2 - 2 k^{2} + 5 k - 2

x_{1, 2} = \frac{- 2 \pm 2 - 2 k ^{2} + 5 k - 2}{2 ( k - 1 )}; k \neq = 1

Trenne die Lösungen

x_{1} = \frac{- 2 + 2 - 2 k ^{2} + 5 k - 2}{2 ( k - 1 )}, x_{2} = \frac{- 2 - 2 - 2 k ^{2} + 5 k - 2}{2 ( k - 1 )}

x = \frac{- 2 + 2 - 2 k ^{2} + 5 k - 2}{2 ( k - 1 )} : \frac{- 1 + - 2 k ^{2} + 5 k - 2}{k - 1}

x = \frac{- 2 - 2 - 2 k ^{2} + 5 k - 2}{2 ( k - 1 )} : - \frac{- 2 k ^{2} + 5 k - 2 + 1}{k - 1}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- 1 + - 2 k ^{2} + 5 k - 2}{k - 1}, x = - \frac{- 2 k ^{2} + 5 k - 2 + 1}{k - 1}; k \neq = 1

a = π \cdot r^{2}

\frac{1}{4} r^{2} + 3 r + 8 = 0

x^{2} + 8 x - 96 = 0

- 26 x + 8 x^{2} = 24

3 x^{2} + 9 x - 7 = 0