To find the value of [tex]\(k\)[/tex] for the quadratic equation [tex]\(3x^2 + 3kx + k = 0\)[/tex], given that one root is twice the other, follow these steps:
### Step 1: Understand the given condition
Assume the roots of the quadratic equation are [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex]. According to the problem, one root is twice the other. Let's denote:
[tex]\[ x_2 = 2x_1 \][/tex]
### Step 2: Use the properties of roots of quadratic equations
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], with roots [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex], the sum and product of the roots are given by:
1. Sum of the roots: [tex]\(x_1 + x_2 = -\frac{b}{a}\)[/tex]
2. Product of the roots: [tex]\(x_1 x_2 = \frac{c}{a}\)[/tex]
### Step 3: Apply the properties to the given equation
For the given equation [tex]\(3x^2 + 3kx + k = 0\)[/tex]:
- Coefficient [tex]\(a = 3\)[/tex]
- Coefficient [tex]\(b = 3k\)[/tex]
- Coefficient [tex]\(c = k\)[/tex]
#### Sum of the roots:
[tex]\[
x_1 + x_2 = -\frac{3k}{3} = -k
\][/tex]
Given that [tex]\(x_2 = 2x_1\)[/tex], we can write:
[tex]\[
x_1 + 2x_1 = -k \quad \Rightarrow \quad 3x_1 = -k \quad \Rightarrow \quad x_1 = -\frac{k}{3}
\][/tex]
#### Product of the roots:
[tex]\[
x_1 x_2 = \frac{k}{3}
\][/tex]
Given that [tex]\(x_2 = 2x_1\)[/tex], we can substitute:
[tex]\[
x_1 (2x_1) = \frac{k}{3} \quad \Rightarrow \quad 2x_1^2 = \frac{k}{3}
\][/tex]
Substitute [tex]\(x_1 = -\frac{k}{3}\)[/tex]:
[tex]\[
2 \left(-\frac{k}{3}\right)^2 = \frac{k}{3} \quad \Rightarrow \quad 2 \cdot \frac{k^2}{9} = \frac{k}{3} \quad \Rightarrow \quad \frac{2k^2}{9} = \frac{k}{3}
\][/tex]
### Step 4: Solve for [tex]\(k\)[/tex]
Multiply both sides by 9 to clear the fraction:
[tex]\[
2k^2 = 3k \quad \Rightarrow \quad 2k^2 - 3k = 0 \quad \Rightarrow \quad k(2k - 3) = 0
\][/tex]
This equation gives us two solutions:
[tex]\[
k = 0 \quad \text{or} \quad 2k - 3 = 0
\][/tex]
Solving [tex]\(2k - 3 = 0\)[/tex]:
[tex]\[
2k = 3 \quad \Rightarrow \quad k = \frac{3}{2}
\][/tex]
### Step 5: Verify the solutions
The possible values for [tex]\(k\)[/tex] satisfying the given condition are:
[tex]\[
k = 0 \quad \text{or} \quad k = \frac{3}{2}
\][/tex]
Thus, the values of [tex]\(k\)[/tex] for which one root of the equation [tex]\(3x^2 + 3kx + k = 0\)[/tex] is twice the other are [tex]\(0\)[/tex] and [tex]\(\frac{3}{2}\)[/tex].
