Find the value of `3a` for which one root of the quadratic equation `(a^2-5a+3)x^2+(3a-1)x+2=0` is twice as large as the other.

Answer

To find the value of \(3a\) for which one root of the quadratic equation \((a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0\) is twice as large as the other, we can follow these steps:

### Step 1: Define the roots
Let the roots of the quadratic equation be \(\alpha\) and \(2\alpha\), where one root is twice the other.

### Step 2: Use the sum of roots
According to Vieta's formulas, the sum of the roots \(\alpha + 2\alpha\) is given by:
\[ \alpha + 2\alpha = 3\alpha = -\frac{b}{a} \]
Here, \(b = 3a - 1\) and \(a = a^2 - 5a + 3\). Thus,
\[ 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \]

### Step 3: Simplify the equation
From the equation above, we can express \(\alpha\):
\[ \alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)} \]

### Step 4: Use the product of roots
The product of the roots \(\alpha \cdot 2\alpha\) is given by:
\[ \alpha \cdot 2\alpha = 2\alpha^2 = \frac{c}{a} \]
Here, \(c = 2\) and \(a = a^2 - 5a + 3\). Thus,
\[ 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \]

### Step 5: Substitute \(\alpha\) into the product equation
Substituting \(\alpha\) from Step 3 into the product equation:
\[ 2\left(\frac{1 - 3a}{3(a^2 - 5a + 3)}\right)^2 = \frac{2}{a^2 - 5a + 3} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2(1 - 3a)^2 = 2 \cdot 9(a^2 - 5a + 3) \]
This simplifies to:
\[ (1 - 3a)^2 = 9(a^2 - 5a + 3) \]

### Step 7: Expand both sides
Expanding both sides:
\[ 1 - 6a + 9a^2 = 9a^2 - 45a + 27 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 1 - 6a + 9a^2 - 9a^2 + 45a - 27 = 0 \]
This simplifies to:
\[ 39a - 26 = 0 \]

### Step 9: Solve for \(a\)
Solving for \(a\):
\[ 39a = 26 \implies a = \frac{26}{39} = \frac{2}{3} \]

### Step 10: Find \(3a\)
Now, we find \(3a\):
\[ 3a = 3 \cdot \frac{2}{3} = 2 \]

Thus, the value of \(3a\) is \(\boxed{2}\).