(2m-3)x^2+(m-3)x-2m+1=0

解

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

x = \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x = \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}

解答ステップ

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( m - 3 ) \pm ( m - 3 ) ^{2} - 4 ( 2 m - 3 ) ( - 2 m + 1 )}{2 ( 2 m - 3 )}

簡素化

(m - 3)^{2} - 4 (2 m - 3) (- 2 m + 1) : 17 m^{2} - 38 m + 21

x_{1, 2} = \frac{- ( m - 3 ) \pm 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}

解を分離する

x_{1} = \frac{- ( m - 3 ) + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x_{2} = \frac{- ( m - 3 ) - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

x = \frac{- ( m - 3 ) + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )} : \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

x = \frac{- ( m - 3 ) - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )} : \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}

二次equationの解:

x = \frac{- m + 3 + 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}, x = \frac{- m + 3 - 17 m ^{2} - 38 m + 21}{2 ( 2 m - 3 )}; m \neq = \frac{3}{2}