Find the value of k such that the sum of the roots of 3x^2+kx-4x=6 is 2

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The value of k has to be determined such that the sum of the roots of 3x^2 + kx - 4x = 6 is 2. As the sum of the roots is 2, let the roots be r and 2 - r

3x^2 + kx - 4x = 6

=> 3x^2 + kx - 4x - 6 = 0

=> x^2 + x(k - 4)/3 - 2 = 0

(x - r)(x - (2 - r)) = x^2 + x(k - 4)/3 - 2

=> x^2 - rx - (2 - r)x + r(2 - r) = x^2 + x(k - 4)/3 - 2

=> x^2 - x(r + 2 - r) + 2r - r^2 = x^2 + x(k - 4)/3 - 2

Equating the coefficients of x, -2 = (k - 4)/3

=> -6 = k - 4

=> k = -2

**The required value of k = -2**

You need to use Vieta's relations such that:

`x_1 + x_2 = (4-k)/3`

You need to remember that the problem provides you the information that the sum of roots is of 2, hence you need to set the equation `(4-k)/3` equal to 2 such that:

`(4-k)/3 = 2`

`4 - k = 6 =gt -k =` `6 - 4`

`k = -2`

**Hence, evaluating k under given conditions yields k = -2.**

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