Question

find k 5x^2-4x+2+k(4x^2-2x-1)=0​

Answer 1

Answer:

[ k € ( - ∞, - 5 / 4 ) U ( - 5 / 4 , - 6 / 5 ) U ( 1 , + ∞ )

k1 = - 6 / 5 , k2 = 1

k € ( - 6 / 5 , 1 ) ]

Step-by-step explanation:

5 x ^ 2 - 4 x + 2 + k ( 4 x ^ 2 - 2 x - 1 ) = 0

5 x ^ 2 - 4 x + 2 + 4 k x ^ 2 - 2 k x - k = 0

5 x ^ 2 - 4 x + 2 + 4 k x ^ 2 - 2 k x - k = 0

( 5 + 4 k ) x ^ 2 - 4 x + 2 - 2 k x - k = 0

( 5 + 4 k ) x ^ 2 - 4 x + 2 - 2 k x - k = 0

( 5 + 4 k ) x ^ 2 + ( - 4 - 2 k ) x + 2 - k = 0

( 5 + 4 k ) x ^ 2 + ( - 4 - 2 k ) x + 2 - k = 0 ,

k ≠ - 5 / 4

( 5 + 4 k ) x ^ 2 + ( - 4 - 2 k ) x + 2 - k = 0 , k ≠ - 5 / 4

D = ( - 4 - 2 k ) ^ 2 - 4 ( 5 + 4 k ) × ( 2 - k ) ,

k ≠ - 5 / 4

D = 20 k ^ 2 + 4 k - 24 , k ≠ - 5 / 4

it's 3 case : D > 0 , D = 0 , D < 0

[ 20 k ^ 2 + 4 k - 24 > 0 , → case 1

20 k ^ 2 + 4 k - 24 = 0 , → case 2

20 k ^ 2 + 4 k - 24 < 0 ] → case 3

[ 20 k ^ 2 + 4 k - 24 > 0 ,

[ 20 k ^ 2 + 4 k - 24 > 0 , 20 k ^ 2 + 4 k - 24 = 0 ,

20 k ^ 2 + 4 k - 24 < 0 ]

20 k ^ 2 + 4 k - 24 > 0 , → case 1

[ k € ( - ∞, - 6 / 5 ) U ( 1 , + ∞ )

20 k ^ 2 + 4 k - 24 = 0 ,

20 k ^ 2 + 4 k - 24 < 0 ]

20 k ^ 2 + 4 k - 24 = 0 , → case 2

k1 = - 6 / 5 , k2 = 1

20 k ^ 2 + 4 k - 24 < 0 ] → case 3

k € ( - 6 / 5 , 1 )

[ k € ( - ∞, - 5 / 4 ) U ( - 5 / 4 , - 6 / 5 ) U ( 1 , + ∞ )

k1 = - 6 / 5 , k2 = 1

k € ( - 6 / 5 , 1 ) ]