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受欢迎的代数 >

(4ka+4a+1)(k+1)=4k^2-7k-11

解答

(4 ka + 4 a + 1) (k + 1) = 4 k^{2} - 7 k - 11

解答

k = - 1, k = - \frac{a + 3}{a - 1}; a \neq = 1

求解步骤

(4 ka + 4 a + 1) (k + 1) = 4 k^{2} - 7 k - 11

展开

(4 ka + 4 a + 1) (k + 1) : 4 a k^{2} + 8 ak + 4 a + k + 1

4 a k^{2} + 8 ak + 4 a + k + 1 = 4 k^{2} - 7 k - 11

将

11

para o lado esquerdo

4 a k^{2} + 8 ak + 4 a + k + 12 = 4 k^{2} - 7 k

将

7 k

para o lado esquerdo

4 a k^{2} + 8 ak + 8 k + 4 a + 12 = 4 k^{2}

将

4 k^{2}

para o lado esquerdo

4 a k^{2} + 8 ak + 8 k + 4 a + 12 - 4 k^{2} = 0

改写成标准形式

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

(4 a - 4) k^{2} + (8 a + 8) k + 4 a + 12 = 0

使用求根公式求解

k_{1, 2} = \frac{- ( 8 a + 8 ) \pm ( 8 a + 8 ) ^{2} - 4 ( 4 a - 4 ) ( 4 a + 12 )}{2 ( 4 a - 4 )}

(8 a + 8)^{2} - 4 (4 a - 4) (4 a + 12) = 16

k_{1, 2} = \frac{- ( 8 a + 8 ) \pm 16}{2 ( 4 a - 4 )}; a \neq = 1

将解分隔开

k_{1} = \frac{- ( 8 a + 8 ) + 16}{2 ( 4 a - 4 )}, k_{2} = \frac{- ( 8 a + 8 ) - 16}{2 ( 4 a - 4 )}

k = \frac{- ( 8 a + 8 ) + 16}{2 ( 4 a - 4 )} : - 1

k = \frac{- ( 8 a + 8 ) - 16}{2 ( 4 a - 4 )} : - \frac{a + 3}{a - 1}

二次方程组的解是:

k = - 1, k = - \frac{a + 3}{a - 1}; a \neq = 1

作图

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