Find all real values of the real parameter $a$ such that the equation \[ 2x^{2}-6ax+4a^{2}-2a-2+\log_{2}(2x^{2}+2x-6ax+4a^{2})= \] \[ =\log_{2}(x^{2}+2x-3ax+2a^{2}+a+1). \] has a unique solution.
Problem
Source: Moldova MO 2006
Tags: parameterization, logarithms, algebra unsolved, algebra
Gabriel(fr)
23.03.2006 01:55
If $2x^{2}+2x-6ax+4a^{2}=S$ and $2(x+a+1)=P$ we can rewrite the equation as,
$S-P + log_2({S})= log_2({\frac{S+P}{2}) \Longleftrightarrow S-P=log_2(\frac{S+P}{2S})}$
but note that,
if $(S-P) > 0 (ie. S>P) \implies \frac{S+P}{2S}>1 \implies S+P>2S \Longleftrightarrow P>S$ (contratidion!)
if $(S-P) < 0 (ie. S<P) \implies \frac{S+P}{2S}<1 \implies S+P<2S \Longleftrightarrow P<S$ (contratidion!)
consequently the only possible values of $P$ and $S$ occur when $P=S$ therefore,
$S-P=log_2({1})=0$ which is true. Now we need to find the value of $x$ such that the equation $S=P$ has a unique solution, consequently,
$2x^{2}+2x-6ax+4a^{2}=2(x+a+1) \implies x^{2}-3ax+2a^2-a-1=0$ this second-degree equation has a unique root when its discriminant is zero then,
$\Delta=\sqrt{a^2+4a+4}=0 \implies a=-2$
Sasha
23.03.2006 21:33
One little mistake: not for both solutions of $x^{2}-3ax+2a^2-a-1=0$ the equation above necessarily makes sense. One can check that for $a\in(-\frac13;0)$ exactly for one of them the under-logarythm expressions are positive. Hence it is a solution too.
mathwizarddude
15.02.2013 09:34
Gabriel(fr) wrote: but note that, if $(S-P) > 0 (ie. S>P) \implies \frac{S+P}{2S}>1 \implies S+P>2S \Longleftrightarrow P>S$ (contratidion!) if $(S-P) < 0 (ie. S<P) \implies \frac{S+P}{2S}<1 \implies S+P<2S \Longleftrightarrow P<S$ (contratidion!) But that's only true when $S>0$.
Rendal
07.09.2018 04:42
Gabriel(fr) wrote:
If $2x^{2}+2x-6ax+4a^{2}=S$ and $2(x+a+1)=P$ we can rewrite the equation as,
$S-P + log_2({S})= log_2({\frac{S+P}{2}) \Longleftrightarrow S-P=log_2(\frac{S+P}{2S})}$
but note that,
if $(S-P) > 0 (ie. S>P) \implies \frac{S+P}{2S}>1 \implies S+P>2S \Longleftrightarrow P>S$ (contratidion!)
if $(S-P) < 0 (ie. S<P) \implies \frac{S+P}{2S}<1 \implies S+P<2S \Longleftrightarrow P<S$ (contratidion!)
consequently the only possible values of $P$ and $S$ occur when $P=S$ therefore,
$S-P=log_2({1})=0$ which is true. Now we need to find the value of $x$ such that the equation $S=P$ has a unique solution, consequently,
$2x^{2}+2x-6ax+4a^{2}=2(x+a+1) \implies x^{2}-3ax+2a^2-a-1=0$ this second-degree equation has a unique root when its discriminant is zero then,
$\Delta=\sqrt{a^2+4a+4}=0 \implies a=-2$
Is that full solution?