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人気のある代数 >

(b-c)*x^2+(c-a)x+(a-b)=0

解

(b - c) \cdot x^{2} + (c - a) x + (a - b) = 0

解

x = \frac{- c + a + c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}, x = \frac{- c + a - c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}; b - c \neq = 0

解答ステップ

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

拡張

(b - c) x^{2} + (c - a) x + (a - b) : b x^{2} - c x^{2} + c x - a x + a - b

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

標準的な形式で書く

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

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

解くとthe二次式

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

簡素化

(c - a)^{2} - 4 (b - c) (a - b) : c^{2} - 4 b c + 2 c a + 4 b^{2} + a^{2} - 4 ba

x_{1, 2} = \frac{- ( c - a ) \pm c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}; b - c \neq = 0

解を分離する

x_{1} = \frac{- ( c - a ) + c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}, x_{2} = \frac{- ( c - a ) - c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}

x = \frac{- ( c - a ) + c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )} : \frac{- c + a + c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}

x = \frac{- ( c - a ) - c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )} : \frac{- c + a - c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}

二次equationの解:

x = \frac{- c + a + c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}, x = \frac{- c + a - c ^{2} - 4 b c + 2 c a + 4 b ^{2} + a ^{2} - 4 ba}{2 ( b - c )}; b - c \neq = 0

グラフ

人気の例

solvefor x,p=(x-1)(x+1)p(2)

so l v e f or x, p = (x - 1) (x + 1) p (2)

x^{2} - 18 x - 77 = 0

4 t (3 - t) = 4 t + 3

(a + c) (a - c) = 0

solvefor x,((p-3)(p+5))*x^2-4x+8=0

so l v e f or x, ((p - 3) (p + 5)) \cdot x^{2} - 4 x + 8 = 0