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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