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Step-by-Step Solution
Step 1: Identify the roots and their relationship
Let the two roots of the quadratic equation
$$\bigl(a^2 - 5a + 3\bigr)x^2 \;+\; \bigl(3a - 1\bigr)x \;+\; 2 \;=\; 0$$
be $ \alpha $ and $2\alpha $. This means one root is twice the other.
Step 2: Write down the sum and product of the roots
For a quadratic equation of the form
$$Ax^2 + Bx + C = 0,$$
the sum of the roots $ (\alpha_1 + \alpha_2) $ is given by
$$-\frac{B}{A},$$
and the product of the roots $ (\alpha_1 \cdot \alpha_2) $ is given by
$$\frac{C}{A}.$$
Step 3: Apply these formulas to the given equation
Here,
$A = a^2 - 5a + 3, \quad B = 3a - 1, \quad C = 2.$
Because our roots are $ \alpha $ and $ 2\alpha $, we have:
Sum of roots:
$$\alpha + 2\alpha = 3\alpha = -\frac{B}{A} = -\frac{3a - 1}{a^2 - 5a + 3} = \frac{1 - 3a}{a^2 - 5a + 3}.$$
Hence,
$$3\alpha = \frac{1 - 3a}{a^2 - 5a + 3} \quad\Rightarrow\quad \alpha = \frac{1 - 3a}{3\bigl(a^2 - 5a + 3\bigr)}.$$
Product of roots:
$$\alpha \cdot (2\alpha) = 2\alpha^2 = \frac{C}{A} = \frac{2}{a^2 - 5a + 3}.$$
Step 4: Express $ \alpha^2 $ and form an equation
From
$$2\alpha^2 = \frac{2}{a^2 - 5a + 3},$$
we get
$$\alpha^2 = \frac{1}{a^2 - 5a + 3}.$$
But from the expression for $ \alpha $ above, we also have
$$\alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)} \quad\Longrightarrow\quad \alpha^2 = \frac{\bigl(1 - 3a\bigr)^2}{9\bigl(a^2 - 5a + 3\bigr)^2}.$$
Step 5: Equate the two expressions for $ \alpha^2 $
Equating the two forms of $ \alpha^2 $:
$$\frac{\bigl(1 - 3a\bigr)^2}{9\bigl(a^2 - 5a + 3\bigr)^2} = \frac{1}{a^2 - 5a + 3}.$$
Multiply both sides by $ 9\bigl(a^2 - 5a + 3\bigr)^2 $ to clear the denominator:
$$\bigl(1 - 3a\bigr)^2 = 9\bigl(a^2 - 5a + 3\bigr).$$
Step 6: Simplify the resulting expression
Expand both sides:
Left side:
$$(1 - 3a)^2 = 1 - 6a + 9a^2.$$
Right side:
$$9(a^2 - 5a + 3) = 9a^2 - 45a + 27.$$
Equate:
$$1 - 6a + 9a^2 = 9a^2 - 45a + 27.$$
Step 7: Solve for $a$
Subtract $ 9a^2 $, $ 1 $, and add $ 6a $ on both sides to simplify:
\[
1 - 6a + 9a^2 - 9a^2 = -45a + 27 - 1 + 6a,
\]
which becomes
\[
-6a + 1 = -45a + 27 + 6a \quad\Rightarrow\quad -6a + 1 = -39a + 27.
\]
Move terms involving $a$ to one side and constants to the other:
\[
-6a + 39a = 27 - 1
\quad\Rightarrow\quad
33a = 26
\quad\Rightarrow\quad
a = \frac{26}{33} = \frac{2}{3}.
\]
Step 8: Conclude the value of $a$
Hence, the value of $a$ for which one root is twice the other is
$$ \frac{2}{3}.$$
