If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( p x^{2}+q x+r=0 \), express \( \left(\left(2+\alpha^{2}\right)\left(2+\beta^{2}\right)\right. \) in terms of \( p \) and \( q \)

To express \( \left( (2 + \alpha^2)(2 + \beta^2) \right) \) in terms of \( p \) and \( q \), we can start by expanding the expression: \[ (2 + \alpha^2)(2 + \beta^2) = 2 \cdot 2 + 2 \cdot \beta^2 + 2 \cdot \alpha^2 + \alpha^2 \beta^2 \] This simplifies to: \[ = 4 + 2(\alpha^2 + \beta^2) + \alpha^2 \beta^2 \] Next, we need to find \( \alpha^2 + \beta^2 \) and \( \alpha^2 \beta^2 \) in terms of \( p \) and \( q \). Using Vieta's formulas, we know: - \( \alpha + \beta = -\frac{q}{p} \) - \( \alpha \beta = \frac{r}{p} \) Now, we can find \( \alpha^2 + \beta^2 \): \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] Substituting the values from Vieta's formulas: \[ \alpha^2 + \beta^2 = \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \] This simplifies to: \[ \alpha^2 + \beta^2 = \frac{q^2}{p^2} - \frac{2r}{p} \] Next, we find \( \alpha^2 \beta^2 \): \[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{r}{p}\right)^2 = \frac{r^2}{p^2} \] Now we can substitute these results back into our expression for \( (2 + \alpha^2)(2 + \beta^2) \): \[ (2 + \alpha^2)(2 + \beta^2) = 4 + 2\left(\frac{q^2}{p^2} - \frac{2r}{p}\right) + \frac{r^2}{p^2} \] This simplifies to: \[ = 4 + \frac{2q^2}{p^2} - \frac{4r}{p} + \frac{r^2}{p^2} \] Combining the terms gives: \[ = 4 + \frac{2q^2 + r^2 - 4pr}{p^2} \] Thus, the final expression for \( (2 + \alpha^2)(2 + \beta^2) \) in terms of \( p \) and \( q \) is: \[ \boxed{4 + \frac{2q^2 + r^2 - 4pr}{p^2}} \]