For which real values of parameter $a$ both roots of polynom $f(x)=(a+1)x^2 + 2ax + a +3$ are positive numbers.
In solution they give 3 conditions that have to be satisfied.
1) $(a+1)f(0)>0$
2) $D>0$
3) $x_{0}>0$
First 2 i understand but not last one.
I dont know what $x_{0}>0$ is . When i calculate first 2 conditions i get
$a\in \left \langle -\infty,-3 \right \rangle \cup \left \langle -1,-3/4 \right \rangle$
But books final solution is only $ \left \langle -1,-3/4 \right \rangle$
The discriminant must be positive:
$$\Delta=4a^2-4(a+1)(a+3)>0\implies-16a-12>0\implies a<-\frac{12}{16}=-\frac34$$
Both roots, say $\;x_1,x_2\;$ , positive:
$$\begin{cases}0<x_1x_2=\cfrac{a+3}{a+1}\iff a<-3\;\;\text{or}\;\;a>-1\\{}\\\text{And}\\{}{}\\ 0<x_1+x_2=-\frac{2a}{a+1}\implies\frac a{a+1}<0\iff -1<a<0\end{cases}$$
Now put things together and do a little mathematics here.
It must be $$-\frac{a}{a+1}+\frac{\sqrt{-4a-3}}{a+1}>0$$ and $$-\frac{a}{a+1}-\frac{\sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$
