# 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?

• Seems to me that according to the rules, $t$ must have a specific, fixed value. – amd May 30 '17 at 6:27
• Are you sure that it’s possible? What effect do the two allowable transformations have on a parabola? – amd May 30 '17 at 6:30
• I am not completely sure ...but if $t$ has a fixed value ...then I believe it maybe impossible.. – user35508 May 30 '17 at 6:32

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 $5$.

Let $f(x)=ax^2+bx+c$. The 2 operations than can be done is equivalent to converting $f(x)$ to the following respectively:

1. $x^2f(\frac{1}{x})$
2. $f(x+t)$

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.

• "(...) each transformation preserve the rationality of the roots (...)" What about the second transformation for say $t=\pi$? – Marc Paul May 30 '17 at 11:17
• @Servaes But that creates a new problem, because the property of rational difference between the roots is not preserved under the first transformation (e.g. if $\alpha = \pi$, $\beta = \pi +1$, then $\alpha - \beta$ is rational and $1/\alpha - 1/\beta$ is not). – Marc Paul May 30 '17 at 12:49