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 zeroes as half of their product, we can use the quadratic formula.
The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the quadratic equation is 3x^2 + (2k+1)x - k - 5 = 0.
The sum of the zeroes is given by:
Sum of zeroes = -b / a
Half of the product is given by:
(1/2) * (product of zeroes) = c / a
Let's calculate these values:
For the equation 3x^2 + (2k+1)x - k - 5 = 0, the coefficients are:
a = 3
b = 2k + 1
c = -k - 5
The sum of the zeroes is -b/a:
(-b) / a = -(2k + 1) / 3
Half of the product is c/a:
(c) / a = (-k - 5) / 3
Since the problem states that the sum of zeroes is half of their product, we can set up the following equation:
-(2k + 1) / 3 = (-k - 5) / 3
Let's solve for k:
Multiply both sides by 3 to get rid of the denominators:
-(2k + 1) = -k - 5
Simplify:
-2k - 1 = -k - 5
Add 1 to both sides:
-2k = -k - 4
Subtract -k from both sides:
-k = -4
Multiply through by -1:
k = 4
Therefore, the value of k that satisfies the condition is k = 4.