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受欢迎的代数 >

solvefor x,f(x-2)=x^2+15

解答

求解

x, f (x - 2) = x^{2} + 15

解答

x = \frac{f + f ^{2} - 4 ( 15 + 2 f )}{2}, x = \frac{f - f ^{2} - 4 ( 15 + 2 f )}{2}

求解步骤

f (x - 2) = x^{2} + 15

展开

f (x - 2) : x f - 2 f

x f - 2 f = x^{2} + 15

交换两边

x^{2} + 15 = x f - 2 f

将

2 f

para o lado esquerdo

x^{2} + 15 + 2 f = x f

将

x f

para o lado esquerdo

x^{2} + 15 + 2 f - x f = 0

改写成标准形式

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

x^{2} - f x + 15 + 2 f = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - f ) \pm ( - f ) ^{2} - 4 \cdot 1 \cdot ( 15 + 2 f )}{2 \cdot 1}

化简

(- f)^{2} - 4 \cdot 1 \cdot (15 + 2 f) : f^{2} - 4 (15 + 2 f)

x_{1, 2} = \frac{- ( - f ) \pm f ^{2} - 4 ( 15 + 2 f )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - f ) + f ^{2} - 4 ( 15 + 2 f )}{2 \cdot 1}, x_{2} = \frac{- ( - f ) - f ^{2} - 4 ( 15 + 2 f )}{2 \cdot 1}

x = \frac{- ( - f ) + f ^{2} - 4 ( 15 + 2 f )}{2 \cdot 1} : \frac{f + f ^{2} - 4 ( 15 + 2 f )}{2}

x = \frac{- ( - f ) - f ^{2} - 4 ( 15 + 2 f )}{2 \cdot 1} : \frac{f - f ^{2} - 4 ( 15 + 2 f )}{2}

二次方程组的解是:

x = \frac{f + f ^{2} - 4 ( 15 + 2 f )}{2}, x = \frac{f - f ^{2} - 4 ( 15 + 2 f )}{2}

作图

流行的例子

solvefor x,(x-2)^2+(y-3)^2+(z+4)^2=0

so l v e f or x, (x - 2)^{2} + (y - 3)^{2} + (z + 4)^{2} = 0

4(n-2)4(n-2)=12

4 (n - 2) 4 (n - 2) = 12

7.15=(0.75+2c)(0.3+2c)

7.15 = (0.75 + 2 c) (0.3 + 2 c)

-x^2+2x+3=x^2-2x+3

- x^{2} + 2 x + 3 = x^{2} - 2 x + 3

x^{2} + 3 x = 26