12. Given that the roots of the equation \( x^{2}-21 x+4=0 \) are \( \alpha^{2} \) and \( \beta^{2} \), where both \( \alpha \) and \( \beta \) are positive, find (i) \( \alpha \beta \). (ii) \( \alpha+\beta \). (iii) an equation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \). 13. Let \( f(x)=x^{3}-5 x^{2}+2 x+8 \) be a polynomial.

To find \(\alpha \beta\), we can use Vieta's formulas. For the equation \(x^2 - 21x + 4 = 0\), the product of the roots \(\alpha^2 \beta^2\) is given by \(\frac{4}{1} = 4\). Thus, \(\alpha \beta = \sqrt{4} = 2\). Next, for \(\alpha + \beta\), we take the sum of the roots given by Vieta’s formulas again. The sum of the roots \(\alpha^2 + \beta^2\) is \(21\). We can relate this to \(\alpha + \beta\) using the identity \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha \beta\). So, substituting the values we get: \((\alpha + \beta)^2 = 21 + 2 \cdot 2 = 25\). Therefore, \(\alpha + \beta = \sqrt{25} = 5\). To find an equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\), we first note that these roots can also be expressed using simple transformations of Vieta’s formulas. The product of the roots is \(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1\) and the sum is \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{21}{2}\). Thus, the equation can be written as \(x^2 - Sx + P = 0\), or \(x^2 - \frac{21}{2}x + 1 = 0\). Multiplying through by 2 gives \(2x^2 - 21x + 2 = 0\). As for \(f(x)\), the polynomial \(f(x) = x^3 - 5 x^2 + 2 x + 8\) has real roots, and you can further explore its behavior using the Rational Root Theorem and synthetic division to factor or find numerically relevant information about its roots. How cool is it that cubic polynomials can have three real roots or a mix of real and complex?! Exploring graphs or using calculus can make this adventure even more thrilling!