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

Answer:

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

The required value of k is -18

Given the quadratic equation [tex]3x^2+2x+k=0[/tex]

Let the roots of the equation be [tex]x_1 \ and \ x_2[/tex]

From the quadratic equatin:

a = 3, b = 3 and c = k

Sum of root = -b/a

x1 + x2 = -3/3

x1 + x2  = -1 ....................... 1

product of roots = c/a

x1x2 = k/3 ....................... 2

Recall that 2x1=−3x2

x1 = -3x2/2 ........... 3

Substitute equation 3 into 1:

x1 + x2  = -1 ....................... 1

-3x2/2 + x2 = -1

-3x2 + 2x2 = -2

-x2 = -2

x2 = 2

From equation 2:

x1x2 = k/3

2x1 = k/3

-3x2 = k/3

-3(2) = k/3

-6 = k/3

k = -18

Hence the required value of k is -18

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