if alpha and beta are the zeroes of the quadratic polynomial f(x) = x² - 3x - 2 , find a quadratic polynomial whose eroes are 1/ 2 alpha + beta and 1/ 2beta+ alpha

The given polynomial is $f\left(x\right) = x^2-3x-2$
Since $\alpha \mathrm{and} \beta$ are the zeroes of the given polynomial. so we have;
sum of the zeroes of the given polynomial = $\alpha +\beta = \frac{-\left(-3\right)}{1} = 3 ...\left(\mathrm{i}\right)$
And product of the zeroes = $\alpha \beta = \frac{-2}{1} = -2 ...\left(\mathrm{ii}\right)$
Now $\frac{1}{2}\alpha +\beta \mathrm{and} \frac{1}{2}\beta +\alpha$ are the zeroes of the required polynomial, so the required polynomial is given by;
$$\begin{gathered} x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{zeroes}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{the} \mathrm{zeroes}\right) \\ = x^2-\left(\frac{1}{2}\alpha +\beta +\frac{1}{2}\beta +\alpha \right)x+\left(\frac{1}{2}\alpha +\beta \right)\left(\frac{1}{2}\beta +\alpha \right) \\ = x^2-\left(\alpha +\beta +\frac{\alpha +\beta }{2}\right)x+\frac{1}{2}\alpha \left(\frac{1}{2}\beta +\alpha \right)+\beta \left(\frac{1}{2}\beta +\alpha \right) \\ = x^2-\left(3+\frac{3}{2}\right)x+\frac{1}{4}\alpha \beta +\frac{1}{2}{\alpha }^2+\frac{1}{2}{\beta }^2+\alpha \beta \{\mathrm{using} (\mathrm{i}\left)\right\} \\ = x^2-\frac{9}{2}x+\frac{1}{4}\times \left(-2\right)+\frac{1}{2}\left({\alpha }^2+{\beta }^2\right)+\left(-2\right) \{\mathrm{using} (\mathrm{ii}\left)\right\} \\ = x^2-\frac{9}{2}x-\frac{1}{2}-2+\frac{1}{2}\left[{\left(\alpha +\beta \right)}^2-2\alpha \beta \right] \\ = x^2-\frac{9}{2}x-\frac{1}{2}-2+\frac{1}{2}\left[3^2-2\times \left(-2\right)\right] \\ = x^2-\frac{9}{2}x-\frac{1}{2}-2+\frac{13}{2} \\ = x^2-\frac{9}{2}x+\frac{13}{2}-\frac{1}{2}-2 \\ = x^2-\frac{9}{2}x+4 \end{gathered}$$