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人気のある代数 >

x^3-3x+2=(x+a)^2(x+b)

解

x^{3} - 3 x + 2 = (x + a)^{2} (x + b)

解

x = \frac{- a ^{2} - 2 ab - 3 + a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}, x = \frac{- a ^{2} - 2 ab - 3 - a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}; 2 a + b \neq = 0

解答ステップ

x^{3} - 3 x + 2 = (x + a)^{2} (x + b)

拡張

(x + a)^{2} (x + b) : x^{3} + b x^{2} + 2 a x^{2} + 2 ab x + a^{2} x + a^{2} b

x^{3} - 3 x + 2 = x^{3} + b x^{2} + 2 a x^{2} + 2 ab x + a^{2} x + a^{2} b

辺を交換する

x^{3} + b x^{2} + 2 a x^{2} + 2 ab x + a^{2} x + a^{2} b = x^{3} - 3 x + 2

2

を左側に移動します

x^{3} + b x^{2} + 2 a x^{2} + 2 ab x + a^{2} x + a^{2} b - 2 = x^{3} - 3 x

3 x

を左側に移動します

x^{3} + b x^{2} + 2 a x^{2} + 2 ab x + a^{2} x + a^{2} b - 2 + 3 x = x^{3}

x^{3}

を左側に移動します

2 a x^{2} + b x^{2} + a^{2} x + 2 ab x + 3 x + a^{2} b - 2 = 0

標準的な形式で書く

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

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

解くとthe二次式

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

簡素化

(a^{2} + 2 ab + 3)^{2} - 4 (2 a + b) (a^{2} b - 2) : a^{4} - 4 a^{3} b + 6 a^{2} + 12 ab + 16 a + 8 b + 9

x_{1, 2} = \frac{- ( a ^{2} + 2 ab + 3 ) \pm a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}; 2 a + b \neq = 0

解を分離する

x_{1} = \frac{- ( a ^{2} + 2 ab + 3 ) + a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}, x_{2} = \frac{- ( a ^{2} + 2 ab + 3 ) - a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}

x = \frac{- ( a ^{2} + 2 ab + 3 ) + a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )} : \frac{- a ^{2} - 2 ab - 3 + a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}

x = \frac{- ( a ^{2} + 2 ab + 3 ) - a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )} : \frac{- a ^{2} - 2 ab - 3 - a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}

二次equationの解:

x = \frac{- a ^{2} - 2 ab - 3 + a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}, x = \frac{- a ^{2} - 2 ab - 3 - a ^{4} - 4 a ^{3} b + 6 a ^{2} + 12 ab + 16 a + 8 b + 9}{2 ( 2 a + b )}; 2 a + b \neq = 0

グラフ

人気の例

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

4 x^{2} - 40 x + 109 = 0

solvefor x,xy^2+4xy+3x=x^2-1

so l v e f or x, x y^{2} + 4 x y + 3 x = x^{2} - 1

6 x^{2} - 5 x = - 4 x + 2

x^{2} - x - 30 = - 2 x