1. **Problem statement:** Given the quadratic equation $x^2 - px + q = 0$ with roots $\alpha$ and $\beta$, find the quadratic equation whose roots are $\alpha^2 + \frac{1}{\beta^2}$ and $\beta^2 + \frac{1}{\alpha^2}$. Then show that if this new equation has roots, it implies $p^2 = 4q$. 2. **Recall the sum and product of roots:** For $x^2 - px + q = 0$, the sum of roots is $\alpha + \beta = p$ and the product is $\alpha \beta = q$. 3. **Express the new roots:** $$\alpha^2 + \frac{1}{\beta^2} = \alpha^2 + \frac{1}{\beta^2}, \quad \beta^2 + \frac{1}{\alpha^2} = \beta^2 + \frac{1}{\alpha^2}$$ 4. **Sum of new roots:** $$S = \left(\alpha^2 + \frac{1}{\beta^2}\right) + \left(\beta^2 + \frac{1}{\alpha^2}\right) = (\alpha^2 + \beta^2) + \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right)$$ 5. **Simplify $\alpha^2 + \beta^2$ using $\alpha + \beta$ and $\alpha \beta$:** $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$$ 6. **Simplify $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$:** $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2} = \frac{p^2 - 2q}{q^2}$$ 7. **Sum of new roots becomes:** $$S = (p^2 - 2q) + \frac{p^2 - 2q}{q^2} = (p^2 - 2q)\left(1 + \frac{1}{q^2}\right) = (p^2 - 2q)\frac{q^2 + 1}{q^2}$$ 8. **Product of new roots:** $$P = \left(\alpha^2 + \frac{1}{\beta^2}\right)\left(\beta^2 + \frac{1}{\alpha^2}\right) = \alpha^2\beta^2 + \alpha^2 \frac{1}{\alpha^2} + \frac{1}{\beta^2} \beta^2 + \frac{1}{\beta^2} \frac{1}{\alpha^2} = q^2 + 1 + 1 + \frac{1}{q^2} = q^2 + 2 + \frac{1}{q^2} = \left(1 + q^2\right)^2 / q^2$$ 9. **Form the quadratic equation with roots $\alpha^2 + \frac{1}{\beta^2}$ and $\beta^2 + \frac{1}{\alpha^2}$:** $$q^2 x^2 - (p^2 - 2q)(1 + q^2) x + (1 + q^2)^2 = 0$$ 10. **Show that if this equation has roots, then $p^2 = 4q$:** The discriminant $\Delta$ of this quadratic is: $$\Delta = [(p^2 - 2q)(1 + q^2)]^2 - 4 q^2 (1 + q^2)^2 = (1 + q^2)^2 \left[(p^2 - 2q)^2 - 4 q^2\right]$$ For real roots, $\Delta \geq 0$, so: $$(p^2 - 2q)^2 - 4 q^2 \geq 0$$ Expanding: $$p^4 - 4 p^2 q + 4 q^2 - 4 q^2 \geq 0 \implies p^4 - 4 p^2 q \geq 0$$ Factor: $$p^2 (p^2 - 4 q) \geq 0$$ Since $p, q$ are nonzero real variables, for equality (roots exist), $$p^2 - 4 q = 0 \implies p^2 = 4 q$$ **Final answers:** - The quadratic equation with roots $\alpha^2 + \frac{1}{\beta^2}$ and $\beta^2 + \frac{1}{\alpha^2}$ is: $$q^2 x^2 - (p^2 - 2q)(1 + q^2) x + (1 + q^2)^2 = 0$$ - The condition for this equation to have roots is: $$p^2 = 4 q$$