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

Lösung

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

x = \frac{k - 1 + 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}, x = \frac{k - 1 - 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}; k \neq = - 1

Schritte zur Lösung

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

Schreibe

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

um:

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

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

Schreibe in der Standard Form

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

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

Löse mit der quadratischen Formel

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

Vereinfache

(- k + 1)^{2} - 4 (k + 1) (- k - 2) : 5 k^{2} + 10 k + 9

x_{1, 2} = \frac{- ( - k + 1 ) \pm 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}; k \neq = - 1

Trenne die Lösungen

x_{1} = \frac{- ( - k + 1 ) + 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}, x_{2} = \frac{- ( - k + 1 ) - 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}

x = \frac{- ( - k + 1 ) + 5 k ^{2} + 10 k + 9}{2 ( k + 1 )} : \frac{k - 1 + 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}

x = \frac{- ( - k + 1 ) - 5 k ^{2} + 10 k + 9}{2 ( k + 1 )} : \frac{k - 1 - 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{k - 1 + 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}, x = \frac{k - 1 - 5 k ^{2} + 10 k + 9}{2 ( k + 1 )}; k \neq = - 1

0 = x^{2} - 36

(x - 3) 2 x = 0

x^{2} + 3 x^{2} = 16

co m pl e t es q u a re x^{2} + 4 x = 10

co m pl e t es q u a re x^{2} + 4 x = 32