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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:
\[
\boxed{2x^2 - 4x + 5 = 0}.
\]

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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.

Find a quadratic equation with roots 1/α and 1/β, where α and β are roots of 5x^2 - 4x + 2 = 0.

Discover how to find a new quadratic equation with roots 1/α and 1/β, given the original quadratic 5x^2 - 4x + 2 = 0. Learn the process of transforming roots using Vieta's formulas and key algebraic concepts. Suitable for students in Grades 10-12.

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