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受欢迎的代数 >

(a+1)x^2+(2a+1)x-a=30

解答

(a + 1) x^{2} + (2 a + 1) x - a = 30

解答

x = \frac{- 2 a - 1 + 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}, x = \frac{- 2 a - 1 - 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}; a \neq = - 1

求解步骤

(a + 1) x^{2} + (2 a + 1) x - a = 30

将

30

para o lado esquerdo

(a + 1) x^{2} + (2 a + 1) x - a - 30 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( 2 a + 1 ) \pm ( 2 a + 1 ) ^{2} - 4 ( a + 1 ) ( - a - 30 )}{2 ( a + 1 )}

化简

(2 a + 1)^{2} - 4 (a + 1) (- a - 30) : 8 a^{2} + 128 a + 121

x_{1, 2} = \frac{- ( 2 a + 1 ) \pm 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}; a \neq = - 1

将解分隔开

x_{1} = \frac{- ( 2 a + 1 ) + 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}, x_{2} = \frac{- ( 2 a + 1 ) - 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}

x = \frac{- ( 2 a + 1 ) + 8 a ^{2} + 128 a + 121}{2 ( a + 1 )} : \frac{- 2 a - 1 + 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}

x = \frac{- ( 2 a + 1 ) - 8 a ^{2} + 128 a + 121}{2 ( a + 1 )} : \frac{- 2 a - 1 - 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}

二次方程组的解是:

x = \frac{- 2 a - 1 + 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}, x = \frac{- 2 a - 1 - 8 a ^{2} + 128 a + 121}{2 ( a + 1 )}; a \neq = - 1

作图

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