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

解

解く

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

解答ステップ

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

解くとthe二次式

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

簡素化

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

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

二次equationの解:

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