Is the transformation possible? I came across this problem which asks to transform $x^2-x-2$ to $x^2-x-1$ ,if possible, using the following rules:
Given a quadratic equation $ax^2+bx+c$ you can :
1)Interchange $a$ and $c$
2)Replace $x$ by $x+t$ where $t$ is a real number.
My approach:
I wrote the equations as 
${(x-\frac12)}^2-\frac94$ and ${(x-\frac12)}^2-\frac54$
Replacing $x$ by $x+t$ in first equation
${(x-\frac12+t)}^2-\frac94$
And setting this equal to ${(x-\frac12)}^2-\frac54$
Simplifying and solving gives a quadratic with variable $t$
Which on applying quadratic formula gives 
$$t=\frac{1-2x\pm \sqrt{4x^2-4x+5}}{2}$$
So, is this correct? If wrong, could somebody help me in the right direction?
 A: Let $f(x)=ax^2+bx+c$. The 2 operations than can be done is equivalent to converting $f(x)$ to the following respectively: 


*

*$x^2f(\frac{1}{x})$

*$f(x+t)$


Thus, if $\alpha$ and $\beta$ are the roots of $f(x)$, the $1^{st}$ transformation gives a polynomial with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ and the $2^{nd}$ transformation gives a polynomial with roots $\alpha - t$ and $\beta -t$.

Now, the general method to solve such problems is to find something which remains invariant. Now, if we look at the roots of $x^2-x-2$ ($2$ and $-1$) and the roots of $x^2-x-1$ ($\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$), we can see that we start with a polynomial with rational roots and end up with a polynomial with irrational roots which is not possible as long as $t$ is rational.

 But this is not possible if $t$ is irrational. If $t$ is irrational, then we can instead show that the discriminant remains invariant in each transformation, which is what Hagen does.
A: We start with a quadratic $ax^2+bx+c$ and use one of the steps to transform it to $a'x^2+b'x+c'$.
In the first case, we have $a'=c,b'=b,c'=a$, in the second case, we have $a'=a, b'=b+2at, c'=c+bt+t^2$.
Note that in the first case, we have $b'^2-4a'c'=b^2-4ac$ and in the second case $b'^2-4a'c'=b^2+4abt+4a^2t^2-4a(c+bt+t^2)=b^2-4ac$.
In other words, neithre of the allowed steps changes the discriminat of our quadratic.
Hence by starting from $x^2-x-2$ with discriminant $9$, we cannot reach $x^2-x-1$ with discriminant $5$.
