solvefor x,kx^2+8x+4=0

解

x, k x^{2} + 8 x + 4 = 0

x = \frac{2 ( - k + 4 - 2 )}{k}, x = - \frac{2 ( - k + 4 + 2 )}{k}; k \neq = 0

解答ステップ

k x^{2} + 8 x + 4 = 0

解くとthe二次式

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 k \cdot 4}{2 k}

簡素化

8^{2} - 4 k \cdot 4 : 4 4 - k

x_{1, 2} = \frac{- 8 \pm 4 4 - k}{2 k}; k \neq = 0

解を分離する

x_{1} = \frac{- 8 + 4 4 - k}{2 k}, x_{2} = \frac{- 8 - 4 4 - k}{2 k}

x = \frac{- 8 + 4 4 - k}{2 k} : \frac{2 ( - k + 4 - 2 )}{k}

x = \frac{- 8 - 4 4 - k}{2 k} : - \frac{2 ( - k + 4 + 2 )}{k}

二次equationの解:

x = \frac{2 ( - k + 4 - 2 )}{k}, x = - \frac{2 ( - k + 4 + 2 )}{k}; k \neq = 0