solvefor x,kx^2-17x+33=0

解

x, k x^{2} - 17 x + 33 = 0

x = \frac{17 + 289 - 132 k}{2 k}, x = \frac{17 - 289 - 132 k}{2 k}; k \neq = 0

解答ステップ

k x^{2} - 17 x + 33 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 k \cdot 33}{2 k}

簡素化

(- 17)^{2} - 4 k \cdot 33 : 289 - 132 k

x_{1, 2} = \frac{- ( - 17 ) \pm 289 - 132 k}{2 k}; k \neq = 0

解を分離する

x_{1} = \frac{- ( - 17 ) + 289 - 132 k}{2 k}, x_{2} = \frac{- ( - 17 ) - 289 - 132 k}{2 k}

x = \frac{- ( - 17 ) + 289 - 132 k}{2 k} : \frac{17 + 289 - 132 k}{2 k}

x = \frac{- ( - 17 ) - 289 - 132 k}{2 k} : \frac{17 - 289 - 132 k}{2 k}

二次equationの解:

x = \frac{17 + 289 - 132 k}{2 k}, x = \frac{17 - 289 - 132 k}{2 k}; k \neq = 0