3x^2 + 2kx + x - 5 = 0
3x^2 + (2k-5)x - 5 = 0
sum = -b/a = (5-2k)/3
product = c/a = -5/3
So, (5-2k)/3 = 1/2 (-5/3)
5-2k = -5/2
10-4k = -5
4k = 15
k = 15/4
f(x) = 3x^2 + 5/2 x - 5
roots are 1/12 (-5±√265)
sum: -5/6
product: -5/3
sum = 1/2 product
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?
1 answer