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受欢迎的代数 >

solvefor x,f=kx^2+(x+1)x+k+2

解答

求解

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

解答

x = \frac{- 1 + 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}, x = \frac{- 1 - 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}; k \neq = - 1

求解步骤

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

展开

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

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

交换两边

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

将

f

para o lado esquerdo

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

改写成标准形式

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

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

使用求根公式求解

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

化简

1^{2} - 4 (k + 1) (k + 2 - f) : 4 k f + 4 f - 4 k^{2} - 12 k - 7

x_{1, 2} = \frac{- 1 \pm 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}; k \neq = - 1

将解分隔开

x_{1} = \frac{- 1 + 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}, x_{2} = \frac{- 1 - 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}

x = \frac{- 1 + 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}; k \neq = - 1

x = \frac{- 1 - 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}; k \neq = - 1

二次方程组的解是:

x = \frac{- 1 + 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}, x = \frac{- 1 - 4 k f + 4 f - 4 k ^{2} - 12 k - 7}{2 ( k + 1 )}; k \neq = - 1

作图

流行的例子

3 x^{2} + 20 x + 33 = 0

5.4x^2+sqrt(3)x-1=0

5.4 x^{2} + 3 x - 1 = 0

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M = x^{2} a + x^{2} b

(2 x)^{2} = 2 x^{2}