(k^2+3)x^2+(k^3+k+1)x+2k+2=0

解答

(k^{2} + 3) \cdot x^{2} + (k^{3} + k + 1) \cdot x + 2 k + 2 = 0

x = \frac{- k ^{3} - k - 1 + k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}, x = \frac{- k ^{3} - k - 1 - k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}; k \neq = 3 i, k \neq = - 3 i

求解步骤

(k^{2} + 3) x^{2} + (k^{3} + k + 1) x + 2 k + 2 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( k ^{3} + k + 1 ) \pm ( k ^{3} + k + 1 ) ^{2} - 4 ( k ^{2} + 3 ) ( 2 k + 2 )}{2 ( k ^{2} + 3 )}

化简

(k^{3} + k + 1)^{2} - 4 (k^{2} + 3) (2 k + 2) : k^{6} + 2 k^{4} - 6 k^{3} - 7 k^{2} - 22 k - 23

x_{1, 2} = \frac{- ( k ^{3} + k + 1 ) \pm k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}; k \neq = 3 i, k \neq = - 3 i

将解分隔开

x_{1} = \frac{- ( k ^{3} + k + 1 ) + k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}, x_{2} = \frac{- ( k ^{3} + k + 1 ) - k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}

x = \frac{- ( k ^{3} + k + 1 ) + k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )} : \frac{- k ^{3} - k - 1 + k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}

x = \frac{- ( k ^{3} + k + 1 ) - k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )} : \frac{- k ^{3} - k - 1 - k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}

二次方程组的解是:

x = \frac{- k ^{3} - k - 1 + k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}, x = \frac{- k ^{3} - k - 1 - k ^{6} + 2 k ^{4} - 6 k ^{3} - 7 k ^{2} - 22 k - 23}{2 ( k ^{2} + 3 )}; k \neq = 3 i, k \neq = - 3 i

(a + b) 2 x^{2} + 6 (a^{2} - b^{2}) x + 9 (a - b) 2 = 0

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

so l v e f or a, 130 r^{2} - a^{2} r^{3} + a^{2} = 0

(3) x = (x^{2} = 9 - 3 = 5)

(2 x - 3)^{2} - (x + 5)^{2} = 0