Here are the rules
(1) If the discriminant = 0.....only one real [double] root
(2) If the discriminant > 0......two real roots
(3) If the discriminant < 0, no real roots
x^2 - kx + 4 = 0
If this has equal roots , the discriminant = 0...so
b^2 - 4ac = 0
(-k)^2 - 4(1)(4) = 0
k^2 - 16 = 0
k^2 = 16 taking both roots....k = 4 or k = -4
Prove this to yourself x^2 - 4x + 4 = 0 → (x - 2)^2 = 0 x = 2 is a [double] root or
x^2 + 4x + 4 = 0 → (x + 2)^2 =0 and x = -2 is a [double root ]
x^2 - 6x + k =0 equal roots....so....
(-6)^2 - 4 (1)k = 0
36 - 4k = 0 add 4k to both sides
36 = 4k divide both sides by 4
k = 9
Proof x^2 - 6x + 9 = 0 → ( x - 3)^2 = 0 → x = 3 is a [double] root
3x^2 -2x + k = 0
So...the discriminant must be < 0
So
(-2)^2 - 4 (3)k < 0
4 - 12k < 0 add 12k to both sides
4 < 12k divide both sides by 12
(1/3) < k → k > 1/3
Prove this......look at the graph here : https://www.desmos.com/calculator/cffskj6a9v
When k = 1/3....we have a [ double] root at x = 1/3
But....when, for instance, k =2/3, notice that we have no real roots
So any k > 1/3 will produce imaginary roots
