z^2-5z+4=0

解

z^{2} - 5 z + 4 = 0

z = 4, z = 1

解答ステップ

z^{2} - 5 z + 4 = 0

解くとthe二次式

z_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

z_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

解を分離する

z_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, z_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

z = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

z = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

二次equationの解:

z = 4, z = 1