Determine the set values of k for which the equation (3-2k) x² + (2k-3) x + 1 = 0 has no real roots

-0.5 < k < 1.5

Step-by-step explanation:

B² - 4AC < 0

(2k-3) ² - 4 (3-2k) (1) < 0

(2k-3) [2k-3 + 4] < 0

(2k-3) (2k+1) < 0

Roots: k = - 1/2, 3/2

-0.5 < k < 1.5

k = (3 x^2 - 3 x + 1) / (2 x^2 - 2 x)

Step-by-step explanation:

Solve for k:

1 + x (2 k - 3) + x^2 (3 - 2 k) = 0

Expand and collect in terms of k:

1 - 3 x + 3 x^2 + k (2 x - 2 x^2) = 0

Multiply both sides by - 1:

-1 + 3 x - 3 x^2 + k (2 x^2 - 2 x) = 0

Subtract - 3 x^2 + 3 x - 1 from both sides:

k (2 x^2 - 2 x) = 3 x^2 - 3 x + 1

Divide both sides by 2 x^2 - 2 x:

Answer: k = (3 x^2 - 3 x + 1) / (2 x^2 - 2 x)