abx^2+(3a+2b)x+6=0

解

ab x^{2} + (3 a + 2 b) x + 6 = 0

x = - \frac{2}{a}, x = - \frac{3}{b}; ab \neq = 0

解答ステップ

ab x^{2} + (3 a + 2 b) x + 6 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 3 a + 2 b ) \pm ( 3 a + 2 b ) ^{2} - 4 ab \cdot 6}{2 ab}

簡素化

(3 a + 2 b)^{2} - 4 ab \cdot 6 : 3 a - 2 b

x_{1, 2} = \frac{- ( 3 a + 2 b ) \pm ( 3 a - 2 b )}{2 ab}; ab \neq = 0

解を分離する

x_{1} = \frac{- ( 3 a + 2 b ) + 3 a - 2 b}{2 ab}, x_{2} = \frac{- ( 3 a + 2 b ) - ( 3 a - 2 b )}{2 ab}

x = \frac{- ( 3 a + 2 b ) + 3 a - 2 b}{2 ab} : - \frac{2}{a}

x = \frac{- ( 3 a + 2 b ) - ( 3 a - 2 b )}{2 ab} : - \frac{3}{b}

二次equationの解:

x = - \frac{2}{a}, x = - \frac{3}{b}; ab \neq = 0