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

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)

To find the value(s) of k such that each function has 2, 1, and 0 roots, we need to use the discriminant of the quadratic equation. The discriminant is given by the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

Let's consider each function one by one:

1. For f(x) = kx^2 - 9x + 6:
The discriminant is Δ = (-9)^2 - 4(k)(6) = 81 - 24k.

To have 2 real roots, the discriminant Δ > 0. Therefore, 81 - 24k > 0.
Solving this inequality, we get: k < 81/24 = 3.375.

To have 1 real root, the discriminant Δ = 0. Therefore, 81 - 24k = 0.
Solving this equation, we get: k = 81/24 = 3.375.

To have 0 real roots, the discriminant Δ < 0. Therefore, 81 - 24k < 0.
Solving this inequality, we get: k > 81/24 = 3.375.

So, the value of k that gives each function 2, 1, and 0 roots is k = 3.375.

2. For f(x) = 3x^2 + kx + 12:
The discriminant is Δ = (k)^2 - 4(3)(12) = k^2 - 144.

To have 2 real roots, the discriminant Δ > 0. Therefore, k^2 - 144 > 0.
This inequality has no restrictions on k.

To have 1 real root, the discriminant Δ = 0. Therefore, k^2 - 144 = 0.
Solving this equation, we get: k = ±12.

To have 0 real roots, the discriminant Δ < 0. Therefore, k^2 - 144 < 0.
This inequality has no restrictions on k.

So, the values of k that give each function 2, 1, and 0 roots are k = ±12.

3. For f(x) = 9x^2 + 3x + k:
The discriminant is Δ = (3)^2 - 4(9)(k) = 9 - 36k.

To have 2 real roots, the discriminant Δ > 0. Therefore, 9 - 36k > 0.
Solving this inequality, we get: k < 9/36 = 0.25.

To have 1 real root, the discriminant Δ = 0. Therefore, 9 - 36k = 0.
Solving this equation, we get: k = 9/36 = 0.25.

To have 0 real roots, the discriminant Δ < 0. Therefore, 9 - 36k < 0.
Solving this inequality, we get: k > 9/36 = 0.25.

So, the value of k that gives each function 2, 1, and 0 roots is k = 0.25.

In summary:
- For f(x) = kx^2 - 9x + 6: k = 3.375.
- For f(x) = 3x^2 + kx + 12: k = ±12.
- For f(x) = 9x^2 + 3x + k: k = 0.25.