for what value of k does each equaton have: 6x^2+4 kx+(k+3)=0

Answer 1

a = 6
b = 4k
c = k+3

x = [ -4k +/- sqrt (16 k^2 - 24(k+3) )/12

x = [ -k/3 +/- (1/12)sqrt (16 k^2 -24 k -72)

the quantity in the square root must be positive to get a real solution so where is it zero?

16 k^2 - 24 k -72 = 0
4 k^2 - 6 k - 18 = 0
2 k^2 - 3 k - 6 = 0
so real solutions for
k< (3-sqrt 57)/4 or k > (3 + sqrt 57)/4
because it is a parabola that is upright (holds water)