Find the value of $a$ such that $f(x)$ has exactly one root $\alpha$ in interval $(1,2)$ and..... Question
Find the value of $a$ such that equation $$f(x)=x^2+(a-3)x+a=0$$ has exactly one root $\alpha$ between the interval $(1,2)$ and $f(x+\alpha)=0$ has exactly one root between the interval $(0,1)$.
Attempt
Discriminant$=0$ for exactly one root.
$F'(x)=0$ where $x$ will lie between $(1,2)$ and hence another restriction on a and $\alpha$. But how will I implement it on second part of the question $f(x+\alpha)$??
Any hints and suggestions are welcome.Its question number 5.

 A: Write $$a = \underbrace{3x-x^2\over +x+1}_{=g(x)} $$
Now we are searching for $a$ if $y=a$ cuts exactly once $g(x)$ at $\alpha$ which is in interval $(1,2)$ and graph $g(x+\alpha)$ in interval $(0,1)$. If we mark $y=x+\alpha$ we see that $-1<y<0$. If we draw a graph of $g$ we see that $y=a$ can not cut graph of $g$ at the same time in $(-1,0)$ and $(1,2)$:
A: There is no solution.
f is a second order polynomial so having only one solution in (1,2) means that either 
(1) $f(x) = (x-\alpha)^2$ or 
(2) $f(2)$ and $f(1)$ have different signs.
(1) implies $a =\alpha = 1$ which is not in (1,2)
(2) $f(2) = 3a-2$ and $f(1) = 2a-2$ so $f(2)f(1) = (3a-2)(2a-2) = 6(a-2/3)(a-1)$ and therefore, $f(2)f(1)<0 <=> 2/3<a<1$
Thus $a$ has to be in $(\frac{2}{3},1)$.
But $αβ=a<1$ and $α>1$ so $β<1<α$ with $β$ being the second root of $f$. So $f(x)>0$ for $x>α$ and $f(x+α)>0$ for $x>0$.
To sum up, 
$f$ has one root in $(1,2)$ $\implies$ $a\in(\frac{2}{3},1)$ $\implies$ $x \mapsto f(x+\alpha)$ has no root in $(0,1)$
