solvefor z,r^2p^2z^2+2rpz+1=0

Lösung

löse nach

z, r^{2} \cdot p^{2} \cdot z^{2} + 2 r p z + 1 = 0

z = - \frac{1}{r p}; r^{2} p^{2} \neq = 0

Schritte zur Lösung

r^{2} p^{2} z^{2} + 2 r p z + 1 = 0

Löse mit der quadratischen Formel

z_{1, 2} = \frac{- 2 r p \pm ( 2 r p ) ^{2} - 4 r ^{2} p ^{2} \cdot 1}{2 r ^{2} p ^{2}}

(2 r p)^{2} - 4 r^{2} p^{2} \cdot 1 = 0

z_{1, 2} = \frac{- 2 r p \pm 0}{2 r ^{2} p ^{2}}; r^{2} p^{2} \neq = 0

z = \frac{- 2 r p}{2 r ^{2} p ^{2}}; r^{2} p^{2} \neq = 0

Vereinfache

\frac{- 2 r p}{2 r ^{2} p ^{2}} : - \frac{1}{r p}

z = - \frac{1}{r p}; r^{2} p^{2} \neq = 0

Die Lösung für die quadratische Gleichung ist:

z = - \frac{1}{r p}; r^{2} p^{2} \neq = 0

x^{2} + 3 x - (a^{2} + a - 2) = 0

(2 a + 3) x^{2} - 3 (a + 1) x + 4 = 0

3 x^{2} - 2 x + 1 = 2 x^{2} - 3 x + 7

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

x^{1} + x^{2} = 5