Key Concepts and Formulas

• Forming a Quadratic Equation from its Roots: A quadratic equation with roots $r_1$ and $r_2$ can be expressed as $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.
• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$, the sum of roots is $x_1 + x_2 = -\frac{b}{a}$ and the product of roots is $x_1 x_2 = \frac{c}{a}$.
• Algebraic Identity: $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$, which can be rearranged to $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

Step-by-Step Solution

1. Identify the Quadratic Equation for $\alpha$ and $\beta$

We are given that $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$. We want to find the quadratic equation that both $\alpha$ and $\beta$ satisfy. We can rearrange either equation to the standard form $ax^2 + bx + c = 0$. Rearranging $\alpha^2 = 5\alpha - 3$, we get:

$\alpha^2 - 5\alpha + 3 = 0$

Similarly, rearranging $\beta^2 = 5\beta - 3$, we get:

$\beta^2 - 5\beta + 3 = 0$

Thus, $\alpha$ and $\beta$ are the roots of the quadratic equation:

$x^2 - 5x + 3 = 0$

Since $\alpha \ne \beta$, these are two distinct roots.

2. Apply Vieta's Formulas to Find $\alpha + \beta$ and $\alpha \beta$

We have the quadratic equation $x^2 - 5x + 3 = 0$. Applying Vieta's formulas, where $a=1$, $b=-5$, and $c=3$:

• Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5$
• Product of roots: $\alpha \beta = \frac{c}{a} = \frac{3}{1} = 3$

3. Define the New Roots and Calculate their Sum

The new quadratic equation has roots $r_1 = \frac{\alpha}{\beta}$ and $r_2 = \frac{\beta}{\alpha}$. We need to find the sum of these roots:

$r_1 + r_2 = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}$

We know that $\alpha + \beta = 5$ and $\alpha \beta = 3$. We need to find $\alpha^2 + \beta^2$. Using the algebraic identity:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$

Substituting the values we found:

$\alpha^2 + \beta^2 = (5)^2 - 2(3) = 25 - 6 = 19$

Now, substitute this back into the sum of the new roots:

$r_1 + r_2 = \frac{19}{3}$

4. Calculate the Product of the New Roots

Now, let's find the product of the new roots:

$r_1 \cdot r_2 = \left(\frac{\alpha}{\beta}\right) \cdot \left(\frac{\beta}{\alpha}\right) = \frac{\alpha \beta}{\alpha \beta} = 1$

5. Form the New Quadratic Equation

Using the general form $x^2 - (r_1 + r_2)x + (r_1 r_2) = 0$ and the values we calculated:

• Sum of new roots ($r_1 + r_2$) = $\frac{19}{3}$
• Product of new roots ($r_1 r_2$) = $1$

Substitute these values into the general equation:

$x^2 - \left(\frac{19}{3}\right)x + 1 = 0$

6. Simplify the Equation to Integer Coefficients

To eliminate the fraction, multiply the entire equation by $3$:

$3 \cdot \left(x^2 - \frac{19}{3}x + 1\right) = 3 \cdot 0$

$3x^2 - 19x + 3 = 0$

Common Mistakes & Tips

• Using the identity for $\alpha^2 + \beta^2$: It's more efficient to use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$ than to solve for $\alpha$ and $\beta$ directly.
• Apply Vieta's Formulas: Vieta's formulas simplify problems involving symmetric expressions of roots.
• Simplify the final equation: Always present the equation with integer coefficients.

Summary

We identified that $\alpha$ and $\beta$ are roots of $x^2 - 5x + 3 = 0$. Using Vieta's formulas, we found $\alpha + \beta = 5$ and $\alpha \beta = 3$. Then, we calculated the sum and product of the new roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$, using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$. Finally, we constructed the new quadratic equation and simplified it to $3x^2 - 19x + 3 = 0$.

The final answer is \boxed{3{x^2} - 19x + 3 = 0}, which corresponds to option (A).