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

Lösung

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

x = \frac{- k - 1 + - 2 k ^{2} + k + 1}{3 k + 1}, x = - \frac{k + - 2 k ^{2} + k + 1 + 1}{3 k + 1}; k \neq = - \frac{1}{3}

Schritte zur Lösung

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

Löse mit der quadratischen Formel

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

Vereinfache

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

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

Trenne die Lösungen

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

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

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

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- k - 1 + - 2 k ^{2} + k + 1}{3 k + 1}, x = - \frac{k + - 2 k ^{2} + k + 1 + 1}{3 k + 1}; k \neq = - \frac{1}{3}

(6 x + 2)^{2} = 16

\frac{( x + 2 ) ^{2} - ( 2 x + 11 )}{3} = 3

\frac{3}{1} + x - x^{2} + \frac{1}{1} - x^{2} + x + \frac{1}{1} - x = 0

\frac{x}{3} = \frac{x ^{2}}{5}

x^{2} + 2 a - 15 = 0