(k-1)x^2-5x+3k-7=0

Lösung

(k - 1) x^{2} - 5 x + 3 k - 7 = 0

x = \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, x = \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

Schritte zur Lösung

(k - 1) x^{2} - 5 x + 3 k - 7 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( k - 1 ) ( 3 k - 7 )}{2 ( k - 1 )}

Vereinfache

(- 5)^{2} - 4 (k - 1) (3 k - 7) : - 12 k^{2} + 40 k - 3

x_{1, 2} = \frac{- ( - 5 ) \pm - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

Trenne die Lösungen

x_{1} = \frac{- ( - 5 ) + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, x_{2} = \frac{- ( - 5 ) - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

x = \frac{- ( - 5 ) + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )} : \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

x = \frac{- ( - 5 ) - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )} : \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, x = \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

(x + 1)^{3} - (x - 2)^{3} = 27

10 - 3 x^{2} = - 2

4 x^{2} - 20 x = - 29

5 (x - 9)^{2} = 20

so l v e f or y, x^{3} + 4 x^{2} + x y^{2} - 4 y^{2} = 0