well b^2-4ac = 81-24k
if that is positive, two real roots
if it is zero, you have one root twice (the parabola just bounces off the x axis at x = -b/2a)
if it is negative there are no REAL roots (although there are two complex conjugate solutions)
Find the value(s) of k so that each function has 2, 1, and 0 roots
f(x)=kx^2-9x+6
f(x)=3x^2+kx+12
f(x)=9x^2+3x+k
I know to use the discriminant but am not sure where to go from there
1 answer
1 answer