Another equation whose roots are symmetric function of α and β
But $\alpha+ \dfrac{1}{\beta}$ is not a symmetric function of $\alpha,\beta$.
What is true, however, is that if the new roots can be written as $\gamma=h(\alpha,\beta), \delta=h(\beta,\alpha)$ for some function $h$, then $\gamma+\delta$ and $\gamma\delta$ are symmetric functions of $\alpha,\beta$, so it is possible to construct the equation having $\gamma,\delta$ as roots using the symmetric functions of $\alpha,\beta\,$ i.e. without solving the original equation.
In this example, without using OP's partial substitution trick:
$\displaystyle \gamma+\delta=\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}=\left(\alpha+\beta\right)\left(1+\frac{1}{\alpha\beta}\right)=(-3)\left(1+\frac{1}{2}\right)=-\frac{9}{2}$
$\displaystyle \gamma\delta=\alpha\beta+1+1+\frac{1}{\beta\alpha}=2+\alpha\beta+\frac{1}{\alpha\beta}=2+2+\frac{1}{2}=\frac{9}{2}$
It follows that the quadratic with roots $\gamma,\delta$ is $x^2 + \dfrac{9}{2}x+\dfrac{9}{2}=0 \iff 2x^2+9x+9=0\,$.
[ EDIT ] $\;$ Following up on comments regarding this part of the original post.
To obtain second equation we could simply assign $x$( variable of second equation)$=\gamma=\alpha+\frac1\beta=\frac{\alpha\beta+1} \beta=\frac{2+1} \beta=\frac3\beta
=>\beta=\frac3 x$
By putting this value of $\beta$ in the $1^{st}$ equation we ibtain the second equation. Using the value of $\delta$ we will obtain the similar equation.
The notation would be less confusing if it used a different variable for the transformation, for example $\color{red}{y} = \gamma = \dots \implies \beta=\frac{3}{y} \implies f\left(\frac{3}{y}\right) = 0 \implies 2y^2+9y+9=0$.
Just the $\gamma=\frac{3}{\beta}$ condition is not enough to derive the equation. For example, it is also true in this case that $\gamma=\frac{3}{2}\alpha$, but substituting $x=\frac{2}{3}y$ in the original equation does not produce the correct result.
The implicit assumption here is that not only $\gamma = \frac{3}{\beta}$ but also $\delta=\frac{3}{\alpha}$, and this is what allows the substitution to work.
A more direct (and general) way to state it would be: let $x_1=\alpha,x_2=\beta$ be the roots of $f(x)=0$, and let $y_1=\gamma=g(x_1), y_2=\delta=g(x_2)$ for some function $g$. If $g$ is invertible, then the equation having $y_1,y_2$ as roots is $f\left(g^{-1}(x)\right)=0\,$.
Best Answer
For a simple approach consider the function $y=x^4+4x^3-8x^2=x^2\cdot\left((x+2)^2-12\right)$ - the intersections with the line $y=-p$ will give the roots of the original. Since this is just a horizontal line in the normal $x,y$ plane, a quick sketch will show that the number of real roots is governed by the relationship of $p$ to the local minima/maxima of the quartic.
The form of the quartic makes this easy to sketch - and the double root at $x=0$ means the cubic you get on differentiating has an obvious root, leaving a quadratic to factor.