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人気のある代数 >

(k-1)x^2+2(k-2)x+(k-1)=0

解

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

解

x = \frac{- k + 2 + - 2 k + 3}{k - 1}, x = \frac{- k + 2 - - 2 k + 3}{k - 1}; k \neq = 1

解答ステップ

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

拡張

(k - 1) x^{2} + 2 (k - 2) x + (k - 1) : k x^{2} - x^{2} + 2 k x - 4 x + k - 1

k x^{2} - x^{2} + 2 k x - 4 x + k - 1 = 0

標準的な形式で書く

a x^{2} + b x + c = 0

(k - 1) x^{2} + (2 k - 4) x + k - 1 = 0

解くとthe二次式

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

簡素化

(2 k - 4)^{2} - 4 (k - 1) (k - 1) : 2 - 2 k + 3

x_{1, 2} = \frac{- ( 2 k - 4 ) \pm 2 - 2 k + 3}{2 ( k - 1 )}; k \neq = 1

解を分離する

x_{1} = \frac{- ( 2 k - 4 ) + 2 - 2 k + 3}{2 ( k - 1 )}, x_{2} = \frac{- ( 2 k - 4 ) - 2 - 2 k + 3}{2 ( k - 1 )}

x = \frac{- ( 2 k - 4 ) + 2 - 2 k + 3}{2 ( k - 1 )} : \frac{- k + 2 + - 2 k + 3}{k - 1}

x = \frac{- ( 2 k - 4 ) - 2 - 2 k + 3}{2 ( k - 1 )} : \frac{- k + 2 - - 2 k + 3}{k - 1}

二次equationの解:

x = \frac{- k + 2 + - 2 k + 3}{k - 1}, x = \frac{- k + 2 - - 2 k + 3}{k - 1}; k \neq = 1

グラフ

人気の例

(4x-1)^2-4(4x-1)+4=0

(4 x - 1)^{2} - 4 (4 x - 1) + 4 = 0

2 x^{2} = 10 x + 5

(x^{2} + x - 6) = 0

3 x^{2} - x + 12 = 0

x^{2} + 5 x - 9 = 5