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受欢迎的代数 >

solvefor x,f(x+4)=x^2+3x

解答

求解

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

解答

x = \frac{- 3 + f + f ^{2} + 10 f + 9}{2}, x = \frac{- 3 + f - f ^{2} + 10 f + 9}{2}

求解步骤

f (x + 4) = x^{2} + 3 x

展开

f (x + 4) : x f + 4 f

x f + 4 f = x^{2} + 3 x

交换两边

x^{2} + 3 x = x f + 4 f

将

4 f

para o lado esquerdo

x^{2} + 3 x - 4 f = x f

将

x f

para o lado esquerdo

x^{2} + 3 x - 4 f - x f = 0

改写成标准形式

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

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

使用求根公式求解

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

化简

(3 - f)^{2} - 4 \cdot 1 \cdot (- 4 f) : f^{2} + 10 f + 9

x_{1, 2} = \frac{- ( 3 - f ) \pm f ^{2} + 10 f + 9}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( 3 - f ) + f ^{2} + 10 f + 9}{2 \cdot 1}, x_{2} = \frac{- ( 3 - f ) - f ^{2} + 10 f + 9}{2 \cdot 1}

x = \frac{- ( 3 - f ) + f ^{2} + 10 f + 9}{2 \cdot 1} : \frac{- 3 + f + f ^{2} + 10 f + 9}{2}

x = \frac{- ( 3 - f ) - f ^{2} + 10 f + 9}{2 \cdot 1} : \frac{- 3 + f - f ^{2} + 10 f + 9}{2}

二次方程组的解是:

x = \frac{- 3 + f + f ^{2} + 10 f + 9}{2}, x = \frac{- 3 + f - f ^{2} + 10 f + 9}{2}

作图

流行的例子

0 = 4.905 t^{2} + 5 t

solvefor a,a^2+1=b(b+1)

so l v e f or a, a^{2} + 1 = b (b + 1)

x^3-x^2=(x-1)^3

x^{3} - x^{2} = (x - 1)^{3}

0=-0.05(x^2-26x-120)

0 = - 0.05 (x^{2} - 26 x - 120)

solvefor q,y=3q^2-18q+r^2-8r+50

so l v e f or q, y = 3 q^{2} - 18 q + r^{2} - 8 r + 50