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受欢迎的代数 >

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

解答

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

解答

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

求解步骤

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

展开

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

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

改写成标准形式

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

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

使用求根公式求解

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

化简

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

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

将解分隔开

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

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

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

二次方程组的解是:

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

作图

流行的例子

p^{2} + 49 = 14 p

solvefor x,f(x-4)=x^2+2x-6

so l v e f or x, f (x - 4) = x^{2} + 2 x - 6

solvefor z,2q^2=z^2*p^2*(1-p^2)

so l v e f or z, 2 q^{2} = z^{2} \cdot p^{2} \cdot (1 - p^{2})

(((4/3)a+5)/3)^2=100

(\frac{( \frac{4}{3} ) a + 5}{3})^{2} = 100

solvefor x,(x-3)^2+(y-6)^2=0

so l v e f or x, (x - 3)^{2} + (y - 6)^{2} = 0