The problem asks to find a quadratic equation whose roots are $\alpha^2$ and $\beta^2$, given that $\alpha$ and $\beta$ are the roots of the equation:

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

Step 1: Recall Vieta's formulas

For the quadratic equation $ax^2 + bx + c = 0$, the sum and product of roots are: $\text{Sum of roots} = -\frac{b}{a}, \quad \text{Product of roots} = \frac{c}{a}.$ Here, $a = 2$, $b = -3$, and $c = -5$. So: $\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}, \quad \alpha\beta = \frac{-5}{2}.$

Step 2: Find the sum and product of $\alpha^2$ and $\beta^2$

The new roots are $\alpha^2$ and $\beta^2$. Using the formulas:

• $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,
• $\alpha^2 \beta^2 = (\alpha \beta)^2$.

Substitute the known values: $\alpha^2 + \beta^2 = \left(\frac{3}{2}\right)^2 - 2\left(\frac{-5}{2}\right) = \frac{9}{4} + 5 = \frac{29}{4}.$ $\alpha^2 \beta^2 = \left(\frac{-5}{2}\right)^2 = \frac{25}{4}.$

Step 3: Form the quadratic equation

The quadratic equation with roots $\alpha^2$ and $\beta^2$ is: $x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0.$ Substitute $\alpha^2 + \beta^2 = \frac{29}{4}$ and $\alpha^2\beta^2 = \frac{25}{4}$: $x^2 - \frac{29}{4}x + \frac{25}{4} = 0.$

Multiply through by 4 to eliminate the fractions: $4x^2 - 29x + 25 = 0.$

Final Answer:

The quadratic equation is: $4x^2 - 29x + 25 = 0.$

Would you like further clarification or more examples? Here are 5 related questions you might find interesting:

1. How do Vieta's formulas apply to higher-degree polynomials?
2. Can we generalize this process for roots that are cubes ($\alpha^3, \beta^3$)?
3. How would the equation change if the roots were $\frac{1}{\alpha}$ and $\frac{1}{\beta}$?
4. What happens if $\alpha$ and $\beta$ are complex roots?
5. How can we verify the solution graphically or numerically?

Tip: Always simplify expressions involving the sum and product of roots systematically to avoid algebraic errors!