solvefor x,kx^2-12x+4=0

解

x, k x^{2} - 12 x + 4 = 0

x = \frac{2 ( - k + 9 + 3 )}{k}, x = \frac{2 ( - - k + 9 + 3 )}{k}; k \neq = 0

解答ステップ

k x^{2} - 12 x + 4 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 k \cdot 4}{2 k}

簡素化

(- 12)^{2} - 4 k \cdot 4 : 4 9 - k

x_{1, 2} = \frac{- ( - 12 ) \pm 4 9 - k}{2 k}; k \neq = 0

解を分離する

x_{1} = \frac{- ( - 12 ) + 4 9 - k}{2 k}, x_{2} = \frac{- ( - 12 ) - 4 9 - k}{2 k}

x = \frac{- ( - 12 ) + 4 9 - k}{2 k} : \frac{2 ( - k + 9 + 3 )}{k}

x = \frac{- ( - 12 ) - 4 9 - k}{2 k} : \frac{2 ( - - k + 9 + 3 )}{k}

二次equationの解:

x = \frac{2 ( - k + 9 + 3 )}{k}, x = \frac{2 ( - - k + 9 + 3 )}{k}; k \neq = 0