Motivation:
Given the roots of the quadratic $2x^2+6x+7=0$ find a quadratic with roots $\alpha^2-1$ and $\beta^2-1$
I was able to solve this problem in two ways:
Method 1:
Sum of the roots $\alpha+\beta=-\frac{b}{a}$
Product of roots $\alpha\beta=\frac{c}{a}$
Hence $\alpha+\beta=-3$ and $\alpha\beta=\frac{7}{2}$
We want an equation with roots $\alpha^2-1$ and $\beta^2-1$
The sum of the roots of the new quadratic will be $\alpha^2-1+\beta^2-1=\alpha^2+\beta^2-2$
The product of the roots of the new quadratic will be $(\alpha^2-1)(\beta^2-1)=\alpha^2\beta^2-(\alpha+\beta)+1$
We are able to compute $\alpha^2+\beta^2$ as it is $(\alpha+\beta)^2-2\alpha\beta$ and so the problem is solved.
Plugging the numbers gives $4u^2+45=0$
Method 2
Let $u=\alpha^2-1\implies\alpha=\sqrt{u+1}$ but we know that $\alpha$ solves the original equation so:
$$\begin{align}2\alpha^2+6\alpha+7&=0\\2(u+1)+6\sqrt{u+1}+7&=0\\\sqrt{u+1}&=\frac{-2u-9}{6}\\u+1&=\frac{1}{36}(-2u-9)^2\\36u+36&=4u^2+36u+81\\0&=4u^2+45\end{align}$$
Question: The first method clearly uses the values of $\alpha$ and $\beta$ but the second seemingly only requires $\alpha$. How is this possible? Sure, one man's $\alpha$ is another's $\beta$ and so you could relabel as the choice of $\alpha$ and $\beta $ is arbitrary. This is believable because of the symmetry involved in the new roots $\alpha^2-1$ looks much like a $\beta^2-1$ but I feel there must be more to this. Supposing one root of the new quadratic was $\alpha^2-1$ but the other was $\beta^3-2\beta$ or something worse? How would the second method know? This leads me to a more fundamental question.
Are there only certain functions of the roots of an old quadratic that can we can find a new quadratic in this way? I suppose we could consider $(x-f(\alpha)(x-g(\beta)=0$ where $f$ and $g$ are the functions of the old roots but then could we alway compute these numerically?
Thanks for taking the time to read this and for any contributions.
Observe the following: Since $\alpha+\beta=-\frac{b}{a}$, you know that $\alpha+\beta=-3$. Above, you considered the case where $u=\alpha^2-1$. Suppose that you have picked a value for $\alpha$ (there are two solutions to the original quadratic and $\alpha$ could be either of them; the Galois action permutes the two roots without changing the base field).
Now, since $\beta=-\alpha-3$, and you know that $\alpha=\sqrt{u+1}$, you know that $\beta=-\alpha-3=-\sqrt{u+1}-3$. Since $2\beta^2+6\beta+7=0$, we know that $2(u+1+6\sqrt{u+1}+9)-6\sqrt{u+1}-18+7=0$. Simplifying, we have that $6\sqrt{u+1}+2u+9=0$. In other words, $$ \sqrt{u+1}=-\frac{2u+9}{6}. $$ This is exactly the same expression that you found above, but starting with $\beta$.
In general, since the Galois action takes $a+b\alpha$ to $a+b\beta$ (where $a$ and $b$ are in $\mathbb{Q}$), we see that the Galois action takes $\alpha^2-1$ to $\beta^2-1$. Therefore, if $\alpha^2-1$ is the root of a polynomial $p(x)$ with coefficients in $\mathbb{Q}$, then so is $\beta^2-1$. More precisely, if $\sigma$ is the Galois action and $p(x)=\sum a_ix^i$, then you know that $p(\alpha)=0$, but $\sigma(p(\alpha))=\sigma(0)=0$, and $\sigma(\sum a_i\alpha^i)=\sum a_i\sigma(\alpha)^i$.
Since $\alpha^2-1$ and $\beta^2-1$ are Galois conjugates, any polynomial (with coefficients in $\mathbb{Q}$) which vanishes at one of them vanishes at both. That is why you only need to know $\alpha$ in the second approach, the $\beta$ comes for free.