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受欢迎的 Geometry >

solvefor x,x^2-x+1-2k(x)^2+4k+1=0

解答

求解

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

解答

x = \frac{1 + 32 k ^{2} - 7}{2 ( 1 - 2 k )}, x = \frac{1 - 32 k ^{2} - 7}{2 ( 1 - 2 k )}; k \neq = \frac{1}{2}

求解步骤

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

展开

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

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

改写成标准形式

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

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

使用求根公式求解

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

化简

(- 1)^{2} - 4 (1 - 2 k) (4 k + 2) : 32 k^{2} - 7

x_{1, 2} = \frac{- ( - 1 ) \pm 32 k ^{2} - 7}{2 ( 1 - 2 k )}; k \neq = \frac{1}{2}

将解分隔开

x_{1} = \frac{- ( - 1 ) + 32 k ^{2} - 7}{2 ( 1 - 2 k )}, x_{2} = \frac{- ( - 1 ) - 32 k ^{2} - 7}{2 ( 1 - 2 k )}

x = \frac{- ( - 1 ) + 32 k ^{2} - 7}{2 ( 1 - 2 k )} : \frac{1 + 32 k ^{2} - 7}{2 ( 1 - 2 k )}

x = \frac{- ( - 1 ) - 32 k ^{2} - 7}{2 ( 1 - 2 k )} : \frac{1 - 32 k ^{2} - 7}{2 ( 1 - 2 k )}

二次方程组的解是:

x = \frac{1 + 32 k ^{2} - 7}{2 ( 1 - 2 k )}, x = \frac{1 - 32 k ^{2} - 7}{2 ( 1 - 2 k )}; k \neq = \frac{1}{2}

作图

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