Help with parametric quadratic equation

Question

We have the equation $x^2+ax+2a+1=0$ which has real roots $x_1$ and $x_2$ and a is a parameter. I need to answer to the following questions:

1. Find all values of a for which $x_1=(a-1)x_2$.

2. For which values of a the roots $x_1$ and $x_2$ are positive?

Answer 1

Hint:

$$\begin{align} x_1+x_2&=-a\\ x_1 x_2 &= 2a+1 \end{align}$$

Answer 2

1) $$x_1x_2=2a+1$$ $$\therefore x_2=\frac{2a+1}{x_1}$$ $$\therefore x_1=(a-1)\left(\frac{2a+1}{x_1}\right)$$ $$\therefore x_1^2=(a-1)(2a+1)$$ LHS is always greater than zero. $$\therefore (a-1)(2a+1)>0$$ From there, you will get the interval of $a$.

2) Just put RHS of $x_1+x_2$ and $x_1x_2$ greater than zero, you wll get the range of $a$.

Also, you have discriminant greater than zero. $$\therefore a^2-4(2a+1)>0$$ $$\therefore |a-4|>2\sqrt5$$

In the end, find the intersection of the range of values of $a$ from the answer and from the fact that discriminant is greater than zero.