2x^2+(4a-b)x-2ab=0

解

2 x^{2} + (4 a - b) x - 2 ab = 0

x = \frac{b}{2}, x = - 2 a

解答ステップ

2 x^{2} + (4 a - b) x - 2 ab = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 4 a - b ) \pm ( 4 a - b ) ^{2} - 4 \cdot 2 ( - 2 ab )}{2 \cdot 2}

簡素化

(4 a - b)^{2} - 4 \cdot 2 (- 2 ab) : 4 a + b

x_{1, 2} = \frac{- ( 4 a - b ) \pm ( 4 a + b )}{2 \cdot 2}

解を分離する

x_{1} = \frac{- ( 4 a - b ) + 4 a + b}{2 \cdot 2}, x_{2} = \frac{- ( 4 a - b ) - ( 4 a + b )}{2 \cdot 2}

x = \frac{- ( 4 a - b ) + 4 a + b}{2 \cdot 2} : \frac{b}{2}

x = \frac{- ( 4 a - b ) - ( 4 a + b )}{2 \cdot 2} : - 2 a

二次equationの解:

x = \frac{b}{2}, x = - 2 a