If \( \alpha \) and \( \beta \) are the roots of \( a x^{2}+b x+c=0 \), find the equations whose roots are \( \begin{array}{ll}\text { a. } \alpha \beta^{-1}, \beta \alpha^{-1} & \text { b. } \alpha^{2}, \beta^{2} \\ \text { d. } \alpha^{3}, \beta^{3} & \text { e. } \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} \\ \text { If the roots of the equation } a x^{2}+c x+c=0 & \text { are in the ratio } m: n \text {, then prove that }(a \alpha+b)^{-1},(a \beta+b\end{array} \)

To find the equations whose roots are transformations of the roots \( \alpha \) and \( \beta \) of the quadratic equation \( ax^2 + bx + c = 0 \), we will use Vieta's formulas, which state that for a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \): - The sum of the roots \( \alpha + \beta = -\frac{b}{a} \) - The product of the roots \( \alpha \beta = \frac{c}{a} \) Now, let's find the required equations for each case: ### a. Roots \( \alpha \beta^{-1}, \beta \alpha^{-1} \) Let \( r_1 = \alpha \beta^{-1} \) and \( r_2 = \beta \alpha^{-1} \). 1. **Sum of the roots**: \[ r_1 + r_2 = \alpha \beta^{-1} + \beta \alpha^{-1} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \] Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \): \[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2} \] Thus, \[ r_1 + r_2 = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c}{a}} = \frac{b^2 - 2ac}{ac} \] 2. **Product of the roots**: \[ r_1 r_2 = \alpha \beta^{-1} \cdot \beta \alpha^{-1} = 1 \] The equation with roots \( r_1 \) and \( r_2 \) is: \[ x^2 - \left(\frac{b^2 - 2ac}{ac}\right)x + 1 = 0 \] ### b. Roots \( \alpha^2, \beta^2 \) Let \( r_1 = \alpha^2 \) and \( r_2 = \beta^2 \). 1. **Sum of the roots**: \[ r_1 + r_2 = \alpha^2 + \beta^2 = \frac{b^2 - 2ac}{a^2} \] 2. **Product of the roots**: \[ r_1 r_2 = \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{c}{a}\right)^2 \] The equation with roots \( r_1 \) and \( r_2 \) is: \[ x^2 - \left(\frac{b^2 - 2ac}{a^2}\right)x + \left(\frac{c}{a}\right)^2 = 0 \] ### d. Roots \( \alpha^3, \beta^3 \) Let \( r_1 = \alpha^3 \) and \( r_2 = \beta^3 \). 1. **Sum of the roots**: \[ r_1 + r_2 = \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = \left(-\frac{b}{a}\right)\left(\frac{b^2 - 2ac}{a^2}\right) = -\frac{b(b^2 - 2ac)}{a^3} \] 2. **Product of the roots**: \[ r_1 r_2 = \alpha^3 \beta^3 = (\alpha \beta)^3 = \left(\frac{c}{a}\right)^3 \] The equation with roots \( r_1 \) and \( r_2 \) is: \[ x^2 + \frac{b(b^2 - 2ac)}{a^3}x + \left(\frac{c}{a}\right)^3 = 0 \] ### e. Roots \( \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} \) Let \( r_1 = \sqrt{\frac{\alpha}{\beta}} \) and \( r_2 = \sqrt{\frac{\beta}{\alpha}} \). 1. **Sum of the roots**: \[ r_1 + r_2 = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}} = \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}} = -\frac{b}{\sqrt{ac}} \] 2. **Product of the roots**: \[ r_1 r_2 = \sqrt{\frac{\alpha}{\beta}} \cdot \sqrt{\frac{\beta}{\alpha}} = 1 \] The equation with roots \( r_1 \) and \( r_2 \) is: \[ x^2 + \frac{b}{\sqrt{ac}}x + 1 = 0 \] ### Additional Problem If the roots of the equation \( ax^2 + cx + c = 0 \) are in the ratio \( m:n \), let the roots be \( r_1 = km \) and \( r_2 = kn \) for some \( k \). Using Vieta's formulas: - \( r_1 + r_2 = km + kn = k(m+n) = -\frac{c}{a} \) - \( r_1 r_2 = km \cdot kn = k^2mn = \frac{c}{a} \) From the first equation, we have \( k = -\frac{c}{a(m+n)} \). Substituting \( k \) into the second equation: \[ \left(-\frac{c}{a(m+n)}\right)^2 mn = \frac{c}{a} \] This leads to: \[ \frac{c^2 mn}{a^2(m+n)^2} = \frac{c}{a} \] Cross-multiplying gives: \[ c^2 mn = ac(m+n)^2 \] This proves the relationship. For the roots \( (a\alpha + b)^{-1} \) and \( (a\beta + b)^{-1} \), we can apply similar reasoning using Vieta's formulas to derive the new equation.