It is known that x1 and x2 are roots of the equation 3x^2+2x+k=0, where 2x1=−3x2. Find k.

k = 8

Step-by-step explanation:

Since the roots of the equation 3x² + 2x + k are x₁ and x₂ and 2x₁ = 3x₂.

By the roots of an equation, ax² + bx + c , where its roots are  x₁ and x₂. It follows that sum of roots are x₁ + x₂ = -b/a and product of roots are x₁x₂ = c/a.

Comparing both equations, a = 3 b = 2 and c = k.

So, x₁ + x₂ = -b/a = -2/3 and x₁x₂ = c/a = k/3

x₁ + x₂ = -2/3 and x₁x₂ = k/3

3x₁ + 3x₂ = -2 (1) and 3x₁x₂ = k  (2)

Since 2x₁ = -3x₂, and -2x₁ = 3x₂substituting this into equations (1) and (2) above, we have

3x₁ + 3x₂ = -2 (1) and 3x₁x₂ = k  (2)

3x₁ + (-2x₁) = -2  and x₁(3x₂) = k

3x₁ - 2x₁ = -2  and x₁(3x₂) = k  

x₁ = -2  and x₁(-2x₁) = k  

x₁ = -2  and -2x₁² = k  

Substituting x₁ = -2 into -2x₁² = k, we have

-2x₁² = k

-2(-2)² = k

2(4) = k

8 = k

So, k = 8