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受欢迎的代数 >

s^2+8s+25+ks^2+8ks+12k=0

解答

s^{2} + 8 s + 25 + k s^{2} + 8 k s + 12 k = 0

解答

s = \frac{- 4 - 4 k + 4 k ^{2} - 5 k - 9}{1 + k}, s = - \frac{4 k + 4 k ^{2} - 5 k - 9 + 4}{k + 1}; k \neq = - 1

求解步骤

s^{2} + 8 s + 25 + k s^{2} + 8 k s + 12 k = 0

改写成标准形式

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

(1 + k) s^{2} + (8 + 8 k) s + 25 + 12 k = 0

使用求根公式求解

s_{1, 2} = \frac{- ( 8 + 8 k ) \pm ( 8 + 8 k ) ^{2} - 4 ( 1 + k ) ( 25 + 12 k )}{2 ( 1 + k )}

化简

(8 + 8 k)^{2} - 4 (1 + k) (25 + 12 k) : 2 4 k^{2} - 5 k - 9

s_{1, 2} = \frac{- ( 8 + 8 k ) \pm 2 4 k ^{2} - 5 k - 9}{2 ( 1 + k )}; k \neq = - 1

将解分隔开

s_{1} = \frac{- ( 8 + 8 k ) + 2 4 k ^{2} - 5 k - 9}{2 ( 1 + k )}, s_{2} = \frac{- ( 8 + 8 k ) - 2 4 k ^{2} - 5 k - 9}{2 ( 1 + k )}

s = \frac{- ( 8 + 8 k ) + 2 4 k ^{2} - 5 k - 9}{2 ( 1 + k )} : \frac{- 4 - 4 k + 4 k ^{2} - 5 k - 9}{1 + k}

s = \frac{- ( 8 + 8 k ) - 2 4 k ^{2} - 5 k - 9}{2 ( 1 + k )} : - \frac{4 k + 4 k ^{2} - 5 k - 9 + 4}{k + 1}

二次方程组的解是:

s = \frac{- 4 - 4 k + 4 k ^{2} - 5 k - 9}{1 + k}, s = - \frac{4 k + 4 k ^{2} - 5 k - 9 + 4}{k + 1}; k \neq = - 1

作图

流行的例子

(2 x + 1) (5 x - 3) = 0

4 a^{2} - 81 = 0

5 x^{2} + 4 = 21 x

5 (x - 2)^{2} + 11 = 6

192 = 20 t + 2 t^{2}