The problem asks to find a quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$, where $\alpha$ and $\beta$ are the roots of the given quadratic equation $5x^2 - 4x + 2 = 0$.

Step-by-step approach:

1. Sum and product of the original roots $\alpha$ and $\beta$:

   • The sum of the roots, $\alpha + \beta$, is given by $-\frac{\text{coefficient of }x}{\text{leading coefficient}}$: $\alpha + \beta = -\frac{-4}{5} = \frac{4}{5}.$
   • The product of the roots, $\alpha \beta$, is given by $\frac{\text{constant term}}{\text{leading coefficient}}$: $\alpha \beta = \frac{2}{5}.$

2. Sum and product of the new roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$:

   • The sum of the new roots is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{4}{5}}{\frac{2}{5}} = 2.$
   • The product of the new roots is $\frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha \beta}$: $\frac{1}{\alpha \beta} = \frac{1}{\frac{2}{5}} = \frac{5}{2}.$

3. Form the new quadratic equation: The quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is given by: $x^2 - \left(\text{sum of roots}\right)x + \left(\text{product of roots}\right) = 0.$ Substituting the values we calculated: $x^2 - 2x + \frac{5}{2} = 0.$

4. Clear the fraction: Multiply the entire equation by 2 to eliminate the fraction: $2x^2 - 4x + 5 = 0.$

Thus, the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is: $\boxed{2x^2 - 4x + 5 = 0}.$

Would you like further details or clarifications? Here are some related questions:

1. How can you find the discriminant of the quadratic equation?
2. What are the graphical implications of the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$?
3. How would the process change for a different quadratic equation?
4. Can you use this method for cubic or higher-degree polynomials?
5. How are the sum and product of roots related to Vieta’s formulas?

Tip: For any quadratic equation, knowing the sum and product of roots can quickly help you find transformed roots without solving the equation again.