consider the equation

`ax^2 + bx + c =0---(1)`

let the roots of this equation be, `alpha,beta`

then we can write the equation as,

`(x-alpha)(x-beta) = 0 `

` ` `x^2 -(alpha+beta)x +(alpha.beta) = 0` ---(2)

equation (1) can...

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consider the equation

`ax^2 + bx + c =0---(1)`

let the roots of this equation be, `alpha,beta`

then we can write the equation as,

`(x-alpha)(x-beta) = 0 `

` ` `x^2 -(alpha+beta)x +(alpha.beta) = 0` ---(2)

equation (1) can be rewritten as,

`x^2 + (b/a)x + (c/a) = 0 ---(3)`

` ` equation (2) and (3) are equivelent. hence,

`(alpha +beta) = -(b/a)---(4)`

now lets consider the equation in the problem

` `

`kx^2 -(2k+6)x - 6 = 0`

` `this equation is equivelent to equation (1) and hence we have obtained,

a = k

b = -(2k +6)

c = -6

Then we can substitute these values in the equation (4)

`(alpha +beta) = -(b/a)---(4)`

`(alpha +beta) = -(-(2k+6)/k)`

`(alpha +beta) =((2k+6)/k)`

in the question it is said that ,

`alpha+beta = 4`

` `hence,

`((2k+6)/k) = 4`

`2k + 6 = 4k`

`6 = 2k`

`k =3`

` `**Therefore the answer for k is 3**

lets substitute the value k in the equation

`kx^2 -(2k+6)x - 6 = 0`

`3x^2 - 12x -6 = 0`

`x^2 -4x -2 =0`

`x^2 -4x = 2`

`(x-2)^2 = 2 +4`

`x-2 = +-sqrt(6)`

`x = 2 +-sqrt(6)`