(k+2)*x^2-13x+4k-10=0

解答

(k + 2) \cdot x^{2} - 13 x + 4 k - 10 = 0

x = \frac{13 + - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}, x = \frac{13 - - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}; k \neq = - 2

求解步骤

(k + 2) x^{2} - 13 x + 4 k - 10 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 ( k + 2 ) ( 4 k - 10 )}{2 ( k + 2 )}

化简

(- 13)^{2} - 4 (k + 2) (4 k - 10) : - 16 k^{2} + 8 k + 249

x_{1, 2} = \frac{- ( - 13 ) \pm - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}; k \neq = - 2

将解分隔开

x_{1} = \frac{- ( - 13 ) + - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}, x_{2} = \frac{- ( - 13 ) - - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}

x = \frac{- ( - 13 ) + - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )} : \frac{13 + - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}

x = \frac{- ( - 13 ) - - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )} : \frac{13 - - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}

二次方程组的解是:

x = \frac{13 + - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}, x = \frac{13 - - 16 k ^{2} + 8 k + 249}{2 ( k + 2 )}; k \neq = - 2

(7 - 2 f)^{2} = 9

so l v e f or x, f = x^{2} - 3 x - 2

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

p^{2} + 49 = 14 p

so l v e f or x, f (x - 4) = x^{2} + 2 x - 6