(k-1)x^2+k^2x+21=0

Lösung

(k - 1) x^{2} + k^{2} x + 21 = 0

x = \frac{- k ^{2} + k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}, x = \frac{- k ^{2} - k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}; k \neq = 1

Schritte zur Lösung

(k - 1) x^{2} + k^{2} x + 21 = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- k ^{2} \pm ( k ^{2} ) ^{2} - 4 ( k - 1 ) \cdot 21}{2 ( k - 1 )}

Vereinfache

(k^{2})^{2} - 4 (k - 1) \cdot 21 : k^{4} - 84 (k - 1)

x_{1, 2} = \frac{- k ^{2} \pm k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}; k \neq = 1

Trenne die Lösungen

x_{1} = \frac{- k ^{2} + k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}, x_{2} = \frac{- k ^{2} - k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}

x = \frac{- k ^{2} + k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}; k \neq = 1

x = \frac{- k ^{2} - k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}; k \neq = 1

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- k ^{2} + k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}, x = \frac{- k ^{2} - k ^{4} - 84 ( k - 1 )}{2 ( k - 1 )}; k \neq = 1

6 x^{2} = 10 + 11 x

so l v e f or x, x^{2} - 16 x = 5 x - 9

3 x^{2} - 18 x + 7 = 0

x^{2} - 10 x + 75 = 0

10 - 3 x = x^{2} - 6 x