What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real? Problem: What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?
Attempt: Solving the equation for $x$ I get $$x_{1,2}=-\frac{a}{6}\pm\sqrt{\frac{13a^2-12}{36}}.$$
For the roots to be real, the discriminant as to be greater than or equal to zero, so it yields the inequality 
$$13a^2-12 \geq 0\Leftrightarrow -\frac{2\sqrt{39}}{13}\leq a\leq\frac{2\sqrt{39}}{13}.$$
Condition number two is that both roots should be positive. How should I think to proceed?

The equation is : $3x^2+ax-(a^2-1)=0$


*

*First condition is which you identified, that the discriminant must be positive.

*Note that  the abscissa of vertex of this parabola is $\dfrac{-b}{2a}$ . For both roots to be positive, this value must be positive

*Furthermore, since the parabola is upward, $f(0) > 0$ implies that both roots must be positive. 
These are the three conditions which yield the sought answer.
