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

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:

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

-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

2x^2+5x+17

your welcome. : )

c. 6tan80

step-by-step explanation: