<s>[INST] On a polynomial with roots in $[1,3]$ that are also of the form $2+\csc\theta$ 
If the roots of the equation $x^2+bx+c=0$ are  $\alpha,\beta$ such that $1\leq \alpha,\beta\leq 3$ and also
given that $(\csc  \theta+2)^2+b(\csc \theta+2)+c=0$ satisfy for two different values of $\csc \theta$
Then value of $|b+c|$ and $\min (x^2+bx+c)$

$\bf{My\; Try::}$ We can write it as $x^2+bx+c = (x-\alpha)(x-\beta)=x^2-(\alpha+\beta )x+\alpha\beta$
So $\alpha +\beta = -b$ and $\alpha \beta = c$
and $(\csc \theta+2)^2+b(\csc \theta +2)+c = (\csc \theta-\alpha')(\csc \theta-\beta')=\csc^2 \theta -(\alpha'+\beta' )\csc \theta+\alpha'\beta'$
So $$\csc^2 \theta+(4+b)\csc \theta+(4+2b+c) = \csc^2 \theta -(\alpha'+\beta' )\csc \theta+\alpha'\beta'$$
So $\alpha '+\beta'=-(4+b)$ and $\alpha'\beta'=(4+2b+c)$
Now how can i solve it after that, Help required, Thanks
 [/INST] Note that $|\csc\theta|\ge1$, and hence $2+\csc\theta$ must lie outside $(1,3)$. Yet two distinct $2+\csc\theta$ values are the roots of the given quadratic, which is given to have roots within $[1,3]$. Hence the roots are 1 and 3 and the quadratic is
$$(x-1)(x-3)=x^2-4x+3=(x-2)^2-1$$
Thus $|b+c|=|-4+3|=1$ and $\min(x^2+bx+c)=-1$.
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