When are we able to find a quadratic with roots that are a function of another quadratic? 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.
 A: 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.
A: This is my solution:
Note that $2x^2+6x+7=0$ is equivalent to $x^2= -3x-\dfrac 72$
So  


*

*$\alpha + \beta = -3$  

*$\alpha \, \beta = \dfrac 72$.  

*$\alpha^2 - 1 = -3\alpha-\dfrac 92$   

*$\beta^2 - 1 = -3\beta-\dfrac 92$   
We compute
\begin{align}
    \hline
    (\alpha^2 - 1) + (\beta^2 - 1)
    &= (-3\alpha-\dfrac 92)+(-3\beta-\dfrac 92)\\
    &= -3(\alpha + \beta) - 9 \\
    &= 0 \\
    \hline
    (\alpha^2 - 1)(\beta^2 - 1)
    &=(-3\alpha-\dfrac 92)(-3\beta-\dfrac 92) \\
    &= 9\alpha\beta+\dfrac{27}{2}(\alpha + \beta) + \dfrac{81}{4}\\
    &= \dfrac{63}{2} -\dfrac{81}{2}+\dfrac{81}{4} \\
    &=  \dfrac{45}{4}\\
    \hline
\end{align}
So the polynomial with roots $\alpha^2-1$ and $\beta^2-1$ is $4x^2 + 45$.
another way to look at your second method
Let $\alpha$ and $\beta$ be the roots of $Ax^2 + Bx + C$.
Let $u = \alpha^2 - 1$
Then $\alpha = \pm \sqrt{u+1}$.
\begin{align}
    Ax^2 + Bx + C &= 0 \\
    A\alpha^2 + B\alpha + C &= 0 \\
    A(u+1) \pm B\sqrt{u+1} + C &= 0 \\
    \pm B\sqrt{u+1} &= -A(u+1) - C \\
    \pm B\sqrt{u+1} &= -Au -(A+C) \\
    B^2u + B &= A^2u^2 +2A(A+C)u + (A+C)^2 \\
    A^2u^2 +(2A^2-B^2+2AC)u + (A+C)^2 - B &= 0
\end{align}
Letting $\beta = v^2 - 1$ would have resulted in
$A^2v^2 +(2A^2-B^2+2AC)v + (A+C)^2 - B = 0$
So it seems that just using $\alpha$ will work.
