(k+3)(2x)^2+(j-x)x+m=0

Lösung

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

x = \frac{- j + j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}, x = \frac{- j - j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}; k \neq = - \frac{11}{4}

Schritte zur Lösung

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

Schreibe

(k + 3) (2 x)^{2} + (j - x) x + m

um:

4 k x^{2} + 11 x^{2} + j x + m

4 k x^{2} + 11 x^{2} + j x + m = 0

Schreibe in der Standard Form

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

(4 k + 11) x^{2} + j x + m = 0

Löse mit der quadratischen Formel

x_{1, 2} = \frac{- j \pm j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}

x = \frac{- j + j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}; k \neq = - \frac{11}{4}

x = \frac{- j - j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}; k \neq = - \frac{11}{4}

Die Lösungen für die quadratische Gleichung sind:

x = \frac{- j + j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}, x = \frac{- j - j ^{2} - 4 ( 4 k + 11 ) m}{2 ( 4 k + 11 )}; k \neq = - \frac{11}{4}

y^{2} - 15 y + 14 = - 5 y

q u a d r a t i c f or m u l a x^{2} + 2 x + 2 = 0

x^{2} - 3 x = x - 4 x^{2}

co m pl e t es q u a re x^{2} + \frac{4}{5} x

k r^{2} + (2 k + 6) r + 16 = 0