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

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

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step-by-step explanation: