solvefor x,kx^2-3kx+9=0

解

x, k x^{2} - 3 k x + 9 = 0

x = \frac{3 k + 3 k ( k - 4 )}{2 k}, x = \frac{3 k - 3 k ( k - 4 )}{2 k}; k \neq = 0

解答ステップ

k x^{2} - 3 k x + 9 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - 3 k ) \pm ( - 3 k ) ^{2} - 4 k \cdot 9}{2 k}

簡素化

(- 3 k)^{2} - 4 k \cdot 9 : 3 k (k - 4)

x_{1, 2} = \frac{- ( - 3 k ) \pm 3 k ( k - 4 )}{2 k}; k \neq = 0

解を分離する

x_{1} = \frac{- ( - 3 k ) + 3 k ( k - 4 )}{2 k}, x_{2} = \frac{- ( - 3 k ) - 3 k ( k - 4 )}{2 k}

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

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

二次equationの解:

x = \frac{3 k + 3 k ( k - 4 )}{2 k}, x = \frac{3 k - 3 k ( k - 4 )}{2 k}; k \neq = 0