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人気のある代数 >

solvefor x,f(x-1)=x^2+2

解

解く

x, f (x - 1) = x^{2} + 2

解

x = \frac{f + f ^{2} - 4 ( 2 + f )}{2}, x = \frac{f - f ^{2} - 4 ( 2 + f )}{2}

解答ステップ

f (x - 1) = x^{2} + 2

拡張

f (x - 1) : x f - f

x f - f = x^{2} + 2

辺を交換する

x^{2} + 2 = x f - f

f

を左側に移動します

x^{2} + 2 + f = x f

x f

を左側に移動します

x^{2} + 2 + f - x f = 0

標準的な形式で書く

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

x^{2} - f x + 2 + f = 0

解くとthe二次式

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

簡素化

(- f)^{2} - 4 \cdot 1 \cdot (2 + f) : f^{2} - 4 (2 + f)

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

解を分離する

x_{1} = \frac{- ( - f ) + f ^{2} - 4 ( 2 + f )}{2 \cdot 1}, x_{2} = \frac{- ( - f ) - f ^{2} - 4 ( 2 + f )}{2 \cdot 1}

x = \frac{- ( - f ) + f ^{2} - 4 ( 2 + f )}{2 \cdot 1} : \frac{f + f ^{2} - 4 ( 2 + f )}{2}

x = \frac{- ( - f ) - f ^{2} - 4 ( 2 + f )}{2 \cdot 1} : \frac{f - f ^{2} - 4 ( 2 + f )}{2}

二次equationの解:

x = \frac{f + f ^{2} - 4 ( 2 + f )}{2}, x = \frac{f - f ^{2} - 4 ( 2 + f )}{2}

グラフ

人気の例

x^{2} + 2 = 2 x + 1

(x)=3x^2-36,c(5)

(x) = 3 x^{2} - 36, c (5)

(3x-4)^2+(3x-4)=0

(3 x - 4)^{2} + (3 x - 4) = 0

3 x^{2} + 25 x = - 50

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