Find the value of k such that the quadritic poly 3x square plus 2 kx plus x minus k minus 5 has the sum of zeroes as half of there product?
To find the value of k such that the quadratic polynomial 3x^2 + 2kx + x - k - 5 has the sum of its zeroes as half of their product, we can use the Vieta's formulas.
Vieta's formulas state that the sum of the roots (zeroes) of a quadratic equation ax^2 + bx + c = 0 is equal to -b/a, and the product of the roots is equal to c/a.
In our equation, a = 3, b = 2k + 1, and c = -k - 5.
The sum of the roots is given as half of the product, so we have:
(-b/a) = (1/2) * (c/a)
Substituting the values, we get:
-(2k + 1)/3 = (1/2) * (-k - 5)/3
Multiply both sides of the equation by 6 to eliminate the fractions:
-2(2k + 1) = (-k - 5)
Simplify the equation:
-4k - 2 = -k - 5
Move all the terms to one side of the equation:
-4k + k = -5 - 2
Simplify further:
-3k = -7
Divide both sides by -3 to solve for k:
k = -7 / -3
Finally, simplify:
k = 7/3
Therefore, the value of k such that the quadratic polynomial has the sum of its zeroes as half of their product is k = 7/3.