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受欢迎的代数 >

solvefor a,f(a+1)=(a+1)^2+3(a+1)-4

解答

求解

a, f (a + 1) = (a + 1)^{2} + 3 (a + 1) - 4

解答

a = \frac{- 5 + f + f ^{2} - 6 f + 25}{2}, a = \frac{- 5 + f - f ^{2} - 6 f + 25}{2}

求解步骤

f (a + 1) = (a + 1)^{2} + 3 (a + 1) - 4

展开

f (a + 1) : a f + f

展开

(a + 1)^{2} + 3 (a + 1) - 4 : a^{2} + 5 a

a f + f = a^{2} + 5 a

交换两边

a^{2} + 5 a = a f + f

将

f

para o lado esquerdo

a^{2} + 5 a - f = a f

将

a f

para o lado esquerdo

a^{2} + 5 a - f - a f = 0

改写成标准形式

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

a^{2} + (5 - f) a - f = 0

使用求根公式求解

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

化简

(5 - f)^{2} - 4 \cdot 1 \cdot (- f) : f^{2} - 6 f + 25

a_{1, 2} = \frac{- ( 5 - f ) \pm f ^{2} - 6 f + 25}{2 \cdot 1}

将解分隔开

a_{1} = \frac{- ( 5 - f ) + f ^{2} - 6 f + 25}{2 \cdot 1}, a_{2} = \frac{- ( 5 - f ) - f ^{2} - 6 f + 25}{2 \cdot 1}

a = \frac{- ( 5 - f ) + f ^{2} - 6 f + 25}{2 \cdot 1} : \frac{- 5 + f + f ^{2} - 6 f + 25}{2}

a = \frac{- ( 5 - f ) - f ^{2} - 6 f + 25}{2 \cdot 1} : \frac{- 5 + f - f ^{2} - 6 f + 25}{2}

二次方程组的解是:

a = \frac{- 5 + f + f ^{2} - 6 f + 25}{2}, a = \frac{- 5 + f - f ^{2} - 6 f + 25}{2}

作图

流行的例子

x^{2} - 22 x + 57 = 0

4 x^{2} - 40 = 0

4 r^{2} - 1 = 0

-5=q(0.01q-0.4)

- 5 = q (0.01 q - 0.4)

因式分解 2x^2+4x=0

f a c t or 2 x^{2} + 4 x = 0