Question

if kx²-(3k+6)x+4k+8=0 has the sum of its roots is equal to three times the product of the roots then find rhe value of k?​

Answer 1

`{ \tt{k {x}^(2) - (3k + 6)x + (4k + 8) = 0}}`

» Sum is (3k + 6)/k

» Product is (4k + 8)/k

☑ From the data in the question:

`{ \tt{sum = 3(product)}} \\ { \tt{ ((3k + 6))/(k) = 3 { \huge \{} ((4k + 8))/(k){ \huge{ \}}} }} \\ \\ { \tt{3k + 6 = 3(4k + 8)}}`

» Factorise the left hand side equation:

`{ \tt{3(k + 2) = 3(4k + 8)}} \\ { \tt{k + 2 = 4k + 8}} \\ { \tt{4k - k = 2 - 8}} \\ { \tt{3k = - 6}} \\ { \tt{k = - (6)/(3) }} \\ \\ { \boxed{ \tt{ \: k = - 2 \: }}}`