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

解

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

x = \frac{k + k ^{2} - 64}{4}, x = \frac{k - k ^{2} - 64}{4}

解答ステップ

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

解くとthe二次式

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

簡素化

(- k)^{2} - 4 \cdot 2 \cdot 8 : k^{2} - 64

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} - 64}{2 \cdot 2}

解を分離する

x_{1} = \frac{- ( - k ) + k ^{2} - 64}{2 \cdot 2}, x_{2} = \frac{- ( - k ) - k ^{2} - 64}{2 \cdot 2}

x = \frac{- ( - k ) + k ^{2} - 64}{2 \cdot 2} : \frac{k + k ^{2} - 64}{4}

x = \frac{- ( - k ) - k ^{2} - 64}{2 \cdot 2} : \frac{k - k ^{2} - 64}{4}

二次equationの解:

x = \frac{k + k ^{2} - 64}{4}, x = \frac{k - k ^{2} - 64}{4}