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受欢迎的代数 >

solvefor x,f(k-1)=5x^2+4x-5

解答

求解

x, f (k - 1) = 5 x^{2} + 4 x - 5

解答

x = \frac{- 2 + 5 f k - 5 f + 29}{5}, x = - \frac{5 f k - 5 f + 29 + 2}{5}

求解步骤

f (k - 1) = 5 x^{2} + 4 x - 5

展开

f (k - 1) : f k - f

f k - f = 5 x^{2} + 4 x - 5

交换两边

5 x^{2} + 4 x - 5 = f k - f

两边减去

f k - f

5 x^{2} + 4 x - 5 - (f k - f) = f k - f - (f k - f)

化简

5 x^{2} + 4 x - 5 - f k + f = 0

使用求根公式求解

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 ( - 5 - f k + f )}{2 \cdot 5}

化简

4^{2} - 4 \cdot 5 (- 5 - f k + f) : 2 5 f k - 5 f + 29

x_{1, 2} = \frac{- 4 \pm 2 5 f k - 5 f + 29}{2 \cdot 5}

将解分隔开

x_{1} = \frac{- 4 + 2 5 f k - 5 f + 29}{2 \cdot 5}, x_{2} = \frac{- 4 - 2 5 f k - 5 f + 29}{2 \cdot 5}

x = \frac{- 4 + 2 5 f k - 5 f + 29}{2 \cdot 5} : \frac{- 2 + 5 f k - 5 f + 29}{5}

x = \frac{- 4 - 2 5 f k - 5 f + 29}{2 \cdot 5} : - \frac{5 f k - 5 f + 29 + 2}{5}

二次方程组的解是:

x = \frac{- 2 + 5 f k - 5 f + 29}{5}, x = - \frac{5 f k - 5 f + 29 + 2}{5}

作图

流行的例子

x^{2} - 8 x + n = 0

3 x^{2} - 15 x + 21 = 0

x^{2} = 529

completesquare 5x^2-20x+17

co m pl e t es q u a re 5 x^{2} - 20 x + 17

2 x^{2} + 32 x + 128 = 0