cjquines0 wrote: The operations below can be applied on any expression of the form \(ax^2+bx+c\). $(\text{I})$ If \(c \neq 0\), replace \(a\) by \(4a-\frac{3}{c}\) and \(c\) by \(\frac{c}{4}\). $(\text{II})$ If \(a \neq 0\), replace \(a\) by \(-\frac{a}{2}\) and \(c\) by \(-2c+\frac{3}{a}\). $(\text{III}_t)$ Replace \(x\) by \(x-t\), where \(t\) is an integer. (Different values of \(t\) can be used.) Is it possible to transform \(x^2-x-6\) into each of the following by applying some sequence of the above operations? $(\text{a})$ \(5x^2+5x-1\) $(\text{b})$ \(x^2+6x+2\) Let $\Delta=b^2-4ac$ I transforms $\Delta\to\Delta+3$ II transforms $\Delta\to\Delta+6$ III keeps $\Delta$ unchanged. So $\boxed{\text{impossible to reach }(a)}$ since $\Delta\to\Delta+20$ and $20\not\equiv 0\pmod 3$ Concerning $(b)$, we have $\Delta\to\Delta+3$ and so exactly one transformation I and as many transformations III and so : One transformation III $x\to x-u$ (with maybe $u=0$) One transformation I One transformation III $x\to x-v$ (with maybe $v=0$) But then we need $a$ unchanged and so middle transformation I must keep $a$ unchanged and so $c=1$ after the first trransformation III Which unfortunately is impossible So $\boxed{\text{impossible to reach }(b)}$

cjquines0 wrote: The operations below can be applied on any expression of the form \(ax^2+bx+c\). $(\text{I})$ If \(c \neq 0\), replace \(a\) by \(4a-\frac{3}{c}\) and \(c\) by \(\frac{c}{4}\). $(\text{II})$ If \(a \neq 0\), replace \(a\) by \(-\frac{a}{2}\) and \(c\) by \(-2c+\frac{3}{a}\). $(\text{III}_t)$ Replace \(x\) by \(x-t\), where \(t\) is an integer. (Different values of \(t\) can be used.) Is it possible to transform \(x^2-x-6\) into each of the following by applying some sequence of the above operations? $(\text{a})$ \(5x^2+5x-1\) $(\text{b})$ \(x^2+6x+2\) reread the problem I think there is a mistake

hmida99 wrote: reread the problem I think there is a mistake reread the solution I think there is no mistake

pco wrote: hmida99 wrote: reread the problem I think there is a mistake reread the solution I think there is no mistake I asked him to reread the problem , because it is too simple to be an olympiad problem