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受欢迎的代数 >

solvefor x,f(x+3)=x^2+3

解答

求解

x, f (x + 3) = x^{2} + 3

解答

x = \frac{f + f ^{2} - 4 ( 3 - 3 f )}{2}, x = \frac{f - f ^{2} - 4 ( 3 - 3 f )}{2}

求解步骤

f (x + 3) = x^{2} + 3

展开

f (x + 3) : x f + 3 f

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

交换两边

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

将

3 f

para o lado esquerdo

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

将

x f

para o lado esquerdo

x^{2} + 3 - 3 f - x f = 0

改写成标准形式

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

x^{2} - f x + 3 - 3 f = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - f ) \pm ( - f ) ^{2} - 4 \cdot 1 \cdot ( 3 - 3 f )}{2 \cdot 1}

化简

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

x_{1, 2} = \frac{- ( - f ) \pm f ^{2} - 4 ( 3 - 3 f )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - f ) + f ^{2} - 4 ( 3 - 3 f )}{2 \cdot 1}, x_{2} = \frac{- ( - f ) - f ^{2} - 4 ( 3 - 3 f )}{2 \cdot 1}

x = \frac{- ( - f ) + f ^{2} - 4 ( 3 - 3 f )}{2 \cdot 1} : \frac{f + f ^{2} - 4 ( 3 - 3 f )}{2}

x = \frac{- ( - f ) - f ^{2} - 4 ( 3 - 3 f )}{2 \cdot 1} : \frac{f - f ^{2} - 4 ( 3 - 3 f )}{2}

二次方程组的解是:

x = \frac{f + f ^{2} - 4 ( 3 - 3 f )}{2}, x = \frac{f - f ^{2} - 4 ( 3 - 3 f )}{2}

作图

流行的例子

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

x^{2} - 2 x + 1 = 2 x + 5

completesquare-2x^2-12x-9

co m pl e t es q u a re - 2 x^{2} - 12 x - 9

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

4 h^{2} - 12 h + 8 = 0