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.

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.
