We are given that $\alpha$ and $\beta$ are the roots of the quadratic equation:

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

Step 1: Use Vieta's Formulas

For the equation $x^2 - 2x + 3 = 0$, by Vieta's relations, the sum and product of the roots $\alpha$ and $\beta$ can be written as:

• Sum of the roots: $\alpha + \beta = 2$

• Product of the roots: $\alpha \beta = 3$

Step 2: Find the New Roots

We are asked to find the equation whose roots are:

$\frac{\alpha^2}{\beta} \quad \text{and} \quad \frac{\beta^2}{\alpha}$

Let the new roots be $r_1 = \frac{\alpha^2}{\beta}$ and $r_2 = \frac{\beta^2}{\alpha}$.

We need to determine the sum and product of these new roots.

Sum of the New Roots:

The sum of the new roots is:

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

To simplify this, find a common denominator:

$r_1 + r_2 = \frac{\alpha^3 + \beta^3}{\alpha \beta}$

Using the identity for the sum of cubes:

$\alpha^3 + \beta^3 = (\alpha + \beta) \left( \alpha^2 - \alpha \beta + \beta^2 \right)$

We know that:

• $\alpha + \beta = 2$
• $\alpha \beta = 3$

Now, calculate $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = 2^2 - 2 \times 3 = 4 - 6 = -2$

So,

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

Thus:

$\alpha^3 + \beta^3 = (\alpha + \beta) \left( \alpha^2 - \alpha \beta + \beta^2 \right) = 2 \times (-5) = -10$

Therefore, the sum of the new roots is:

$r_1 + r_2 = \frac{-10}{3}$

Product of the New Roots:

The product of the new roots is:

$r_1 \times r_2 = \frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \frac{\alpha \beta}{\alpha \beta} = 1$

Step 3: Write the Equation

The general form of a quadratic equation with roots $r_1$ and $r_2$ is:

$x^2 - (r_1 + r_2)x + r_1r_2 = 0$

Substitute the sum and product we found:

$x^2 - \left( \frac{-10}{3} \right)x + 1 = 0$

Multiply through by 3 to clear the denominator:

$3x^2 + 10x + 3 = 0$

Thus, the required quadratic equation is:

$\boxed{3x^2 + 10x + 3 = 0}$

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Would you like any further clarifications or additional steps explained?

Here are 5 related questions that you might find useful:

1. How do you use Vieta's formulas for higher-degree polynomials?
2. How do you simplify expressions involving roots in quadratic equations?
3. What is the significance of the sum and product of the roots of a quadratic equation?
4. How can we generalize the approach to finding equations for any given transformations of the roots?
5. How can you use the sum of cubes identity in other types of algebraic manipulations?

Tip: Remember that simplifying expressions for roots can often be easier by leveraging known identities and relationships, such as the sum of cubes, rather than trying to solve for the individual roots directly.