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受欢迎的代数 >

solvefor a,S=2pia^2+2piah

解答

求解

a, S = 2 π a^{2} + 2 πah

解答

a = \frac{- πh + π ( π h ^{2} + 2 S )}{2 π}, a = - \frac{πh + π ( π h ^{2} + 2 S )}{2 π}

求解步骤

S = 2 π a^{2} + 2 πah

交换两边

2 π a^{2} + 2 πah = S

将

S

para o lado esquerdo

2 π a^{2} + 2 πah - S = 0

改写成标准形式

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

2 π a^{2} + 2 πha - S = 0

使用求根公式求解

a_{1, 2} = \frac{- 2 πh \pm ( 2 πh ) ^{2} - 4 \cdot 2 π ( - S )}{2 \cdot 2 π}

化简

(2 πh)^{2} - 4 \cdot 2 π (- S) : 2 π (π h^{2} + 2 S)

a_{1, 2} = \frac{- 2 πh \pm 2 π ( π h ^{2} + 2 S )}{2 \cdot 2 π}

将解分隔开

a_{1} = \frac{- 2 πh + 2 π ( π h ^{2} + 2 S )}{2 \cdot 2 π}, a_{2} = \frac{- 2 πh - 2 π ( π h ^{2} + 2 S )}{2 \cdot 2 π}

a = \frac{- 2 πh + 2 π ( π h ^{2} + 2 S )}{2 \cdot 2 π} : \frac{- πh + π ( π h ^{2} + 2 S )}{2 π}

a = \frac{- 2 πh - 2 π ( π h ^{2} + 2 S )}{2 \cdot 2 π} : - \frac{πh + π ( π h ^{2} + 2 S )}{2 π}

二次方程组的解是:

a = \frac{- πh + π ( π h ^{2} + 2 S )}{2 π}, a = - \frac{πh + π ( π h ^{2} + 2 S )}{2 π}

作图

流行的例子

12 x^{2} + 5 x - 4 = 0

0 = - x^{2} + 100 x

4 x^{2} - 9 = 2 x^{2} + 41

- 4 x^{2} + 16 x + 5 = 0

5 u^{2} + u = 4