Key Concepts and Formulas

• For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots $\alpha + \beta = -\frac{B}{A}$.
• For a quadratic equation $Ax^2 + Bx + C = 0$, the product of the roots $\alpha\beta = \frac{C}{A}$.

Step-by-Step Solution

Step 1: Identify Coefficients

The given quadratic equation is $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$. We identify the coefficients:

• $A = a^2 - 5a + 3$
• $B = 3a - 1$
• $C = 2$

Explanation: We are identifying the coefficients A, B, and C from the general quadratic equation form, which will be used in Vieta's formulas.*

Step 2: Define Roots

Let one root be $\alpha$. Since the other root is twice as large, the other root is $2\alpha$.

Explanation: We are setting up the problem by defining the roots based on the problem statement.*

Step 3: Apply Sum of Roots Formula

Using Vieta's formula for the sum of roots: $\alpha + 2\alpha = -\frac{B}{A}$ $3\alpha = -\frac{3a - 1}{a^2 - 5a + 3}$ $3\alpha = \frac{1 - 3a}{a^2 - 5a + 3} \quad (*)$

Explanation: We apply the formula for the sum of the roots, substituting the expressions for A and B and simplifying.*

Step 4: Apply Product of Roots Formula

Using Vieta's formula for the product of roots: $\alpha(2\alpha) = \frac{C}{A}$ $2\alpha^2 = \frac{2}{a^2 - 5a + 3} \quad (**)$

Explanation: We apply the formula for the product of the roots, substituting the expressions for A and C.*

Step 5: Simplify the Product of Roots Equation From equation $(**)$, we can isolate $\alpha^2$: $2\alpha^2 = \frac{2}{a^2 - 5a + 3}$ $\alpha^2 = \frac{1}{a^2 - 5a + 3} \quad (***)$

Explanation: We simplify the product of roots equation by dividing both sides by 2.*

Step 6: Square the Sum of Roots Equation

Square equation $(*)$: $(3\alpha)^2 = \left(\frac{1 - 3a}{a^2 - 5a + 3}\right)^2$ $9\alpha^2 = \frac{(1 - 3a)^2}{(a^2 - 5a + 3)^2} \quad (****)$

Explanation: We square the sum of roots equation to get an expression for $\alpha^2$, which we can then substitute.*

Step 7: Substitute for $\alpha^2$

Substitute $\alpha^2 = \frac{1}{a^2 - 5a + 3}$ (from equation $(***)$) into equation $(****)$: $9\left(\frac{1}{a^2 - 5a + 3}\right) = \frac{(1 - 3a)^2}{(a^2 - 5a + 3)^2}$

Explanation: We substitute the expression for $\alpha^2$ from the product of roots equation into the squared sum of roots equation.*

Step 8: Simplify and Solve for $a$

Multiply both sides by $(a^2 - 5a + 3)^2$: $9(a^2 - 5a + 3) = (1 - 3a)^2$ Expand both sides: $9a^2 - 45a + 27 = 1 - 6a + 9a^2$ Simplify: $9a^2 - 45a + 27 - 9a^2 + 6a - 1 = 0$ $-39a + 26 = 0$ $39a = 26$ $a = \frac{26}{39} = \frac{2}{3}$

Explanation: We simplify the equation by cancelling terms, expanding, and isolating $a$.*

Step 9: Verify $a^2 - 5a + 3 \neq 0$

Check if $a = \frac{2}{3}$ makes the denominator zero: $(\frac{2}{3})^2 - 5(\frac{2}{3}) + 3 = \frac{4}{9} - \frac{10}{3} + 3 = \frac{4 - 30 + 27}{9} = \frac{1}{9} \neq 0$

Since $A \neq 0$, $a = \frac{2}{3}$ is a valid solution.

Explanation: We need to make sure that the leading coefficient is not zero, so that the equation remains quadratic.*

Step 10: Re-examine the problem to obtain the correct answer

Let $\alpha$ and $2\alpha$ be the roots. Then $\alpha+2\alpha = 3\alpha = \frac{-(3a-1)}{a^2-5a+3}$ $\alpha(2\alpha)=2\alpha^2 = \frac{2}{a^2-5a+3}$ $\alpha = \frac{1-3a}{3(a^2-5a+3)}$ $\alpha^2 = \frac{1}{a^2-5a+3}$ Substituting: $\frac{(1-3a)^2}{9(a^2-5a+3)^2} = \frac{1}{a^2-5a+3}$ $(1-3a)^2 = 9(a^2-5a+3)$ $1-6a+9a^2 = 9a^2-45a+27$ $39a = 26$ $a = 2/3$ However, the given answer is $a=-1/3$. Let us try to work backwards.

If $a=-1/3$, then $3a-1 = -1-1 = -2$. $a^2-5a+3 = 1/9+5/3+3 = (1+15+27)/9 = 43/9$. Then $3\alpha = \frac{2}{43/9} = 18/43$, so $\alpha = 6/43$. Then $2\alpha^2 = 2(36/43^2) = 72/43^2 = \frac{2}{43/9} = 18/43$. $72/43^2 = 18/43$ implies $72 = 18(43)$, which is false. Let $2\alpha = k \alpha$. Then $(1+k)\alpha = \frac{1-3a}{a^2-5a+3}$ and $k\alpha^2 = \frac{2}{a^2-5a+3}$ $\alpha = \frac{1-3a}{(1+k)(a^2-5a+3)}$ $\alpha^2 = \frac{2}{k(a^2-5a+3)}$ So $\frac{(1-3a)^2}{(1+k)^2(a^2-5a+3)^2} = \frac{2}{k(a^2-5a+3)}$ $(1-3a)^2k = 2(1+k)^2(a^2-5a+3)$ If $k=2$, we have $2(1-3a)^2 = 18(a^2-5a+3)$ $(1-3a)^2 = 9(a^2-5a+3)$ $1-6a+9a^2 = 9a^2-45a+27$ $39a = 26$ $a = 2/3$ The correct answer must be $a=-1/3$. Let's solve for this. $(3a-1) = -2$ $a^2-5a+3 = 1/9+5/3+3 = (1+15+27)/9 = 43/9$ $3\alpha = 2/(43/9) = 18/43$, so $\alpha = 6/43$. $2\alpha^2 = 2/(43/9)$ $2\alpha^2 = 18/43$, so $\alpha^2 = 9/43$ $\alpha = 3/\sqrt{43}$. Contradiction. If $a=-1/3$, then $(a^2-5a+3)x^2 + (3a-1)x + 2 = 0$ $(1/9+5/3+3)x^2 + (-1-1)x + 2 = 0$ $(1+15+27)x^2/9 -2x + 2=0$ $43/9 x^2 - 2x + 2 = 0$ $43x^2 - 18x + 18 = 0$ $x = \frac{18 \pm \sqrt{18^2-4(43)(18)}}{86}$ $324-4(43)(18) = 324-3096 = -2772$ The discriminant is negative. So $a=-1/3$ is not correct. The question is flawed.

If one root is double the other, then $a=2/3$. Let's check this. $(a^2-5a+3)x^2 + (3a-1)x + 2 = 0$ $(4/9-10/3+3)x^2 + (2-1)x + 2 = 0$ $(4-30+27)x^2/9 + x + 2 = 0$ $(1/9)x^2 + x + 2 = 0$ $x^2+9x+18 = 0$ $(x+3)(x+6) = 0$ $x=-3,-6$. $-6 = 2(-3)$. So one root is double the other.

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when squaring and expanding expressions.
• Always check if the value of $a$ obtained makes the coefficient of $x^2$ equal to zero.
• Double-check your calculations to avoid arithmetic errors.

Summary

By applying Vieta's formulas and the given condition that one root is twice the other, we derived an equation in terms of $a$. Solving this equation, we found $a = \frac{2}{3}$. We also verified that this value of $a$ does not make the leading coefficient zero. Based on the calculations, the correct answer should be 2/3, however, the problem states that the correct answer is A which corresponds to -1/3. There is an error in the problem statement.

The final answer is \boxed{2/3}, which does not correspond to any of the given options.