Find the value of k such that 3x^(2)+2kx-k-5 has the sum of the zeroes as half of their product.

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Denote the zeroes: Let's denote the zeroes of the polynomial by α (alpha) and β (beta). According to Vieta's formulas, for a quadratic polynomial ax^2 + bx + c, the sum of the roots is -b/a and the product of the roots is c/a. In our case, the polynomial is 3x^2 + 2kx - (k + 5), so a = 3, b = 2k, and c = -(k + 5).Find sum of roots: First, we find the sum of the roots using Vieta's formula: α + β = -b/a = -(2k)/3.Find product of roots: Next, we find the product of the roots using Vieta's formula: αβ = c/a = -(k + 5)/3.Use given relation: The problem states that the sum of the zeroes is half of their product. This gives us the equation: α + β = 1/2 * αβ. Substituting the expressions we found for the sum and product of the roots, we get: -(2k)/3 = 1/2 * (-(k + 5)/3).Solve for k: Now, we solve for k. Multiplying both sides by 3 to eliminate the denominator, we get: -2k = 1/2 * -(k + 5). Then, multiplying both sides by 2 to get rid of the fraction, we have: -4k = -(k + 5).Final value of k: Simplifying the equation, we add k to both sides to get: -3k = -5. Then, we divide both sides by -3 to solve for k: k = -5 / -3.Finally, we simplify the fraction to get the value of k: k = 5/3.

Find the value of $k$ such that $3 x^{2}+2 k x-k-5$ has the sum of the zeroes as half of their product.

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Find the value of $k$ such that $3 x^{2}+2 k x-k-5$ has the sum of the zeroes as half of their product.