(2k+2)x^2+(4-4k)x+k-2=0

解

(2 k + 2) x^{2} + (4 - 4 k) x + k - 2 = 0

x = \frac{- 4 + 4 k + 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}, x = \frac{- 4 + 4 k - 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}; k \neq = - 1

解答ステップ

(2 k + 2) x^{2} + (4 - 4 k) x + k - 2 = 0

解くとthe二次式

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

簡素化

(4 - 4 k)^{2} - 4 (2 k + 2) (k - 2) : 8 k^{2} - 24 k + 32

x_{1, 2} = \frac{- ( 4 - 4 k ) \pm 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}; k \neq = - 1

解を分離する

x_{1} = \frac{- ( 4 - 4 k ) + 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}, x_{2} = \frac{- ( 4 - 4 k ) - 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}

x = \frac{- ( 4 - 4 k ) + 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )} : \frac{- 4 + 4 k + 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}

x = \frac{- ( 4 - 4 k ) - 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )} : \frac{- 4 + 4 k - 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}

二次equationの解:

x = \frac{- 4 + 4 k + 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}, x = \frac{- 4 + 4 k - 8 k ^{2} - 24 k + 32}{2 ( 2 k + 2 )}; k \neq = - 1