Finding condition if a quadratic equation have at least one real root given interval of x 
Question: Find value of $a$ if equation $ax^2 + 2x -1 = 0$ has at least one real root for  $3 \leq x < 4$

I know if the restriction of $x$ is not given, I can use $D \geq 0$ to get condition for at least one real root, but if $x$ is restricted then how do I do it?
 A: Consider the values of the quadratic at end points, viz. $9a+5$ and $16a+7$. There is exactly one root in the interval iff $(9a+5)(16a+7)\le 0 \implies a\in [-\frac59, -\frac7{16})$. [Note we can’t have simultaneously both end points zero for any value of $a$.]
The remaining possibility is two roots in the interval, which needs a turning point in the interval and the value at the turning point to have opposite sign to values at end points. The turning point is $x=-1/a\in (3,4) \implies a\in (-\frac13,-\frac14)$ and so the value at the turning point is $1/a-2/a-1=-1/a-1>0$. However it is seen $9a+5$ (or $16a+7$) is positive as well for allowable $a$, so there cannot be two roots in the interval. 
A: From the quadratic equation,
$$a=\frac{1-2x}{x^2}$$ and the range of this expression for $3\le x<4$ is
$$-\frac59\le a<-\frac7{16}.$$
A: Write the two roots and place them between $3$ and $4$ and then, first suppose $a$ is positive and multiple the three side of the inequalities by $a$ and solve them!. When $a$ is negative the process is like before.
A: $$ax^2 + 2x -1 = 0 \iff x^2 +\dfrac 2a - \dfrac 1a=0$$
I see that there is a much simpler solution, But I liked the way this solution worked out.
Let the roots be $r$ and $s$. Then
$$r + s = -\dfrac 2a \quad \text{and} \quad rs = -\dfrac 1a$$
$$r+s = 2rs$$
$$r(1-2s) = -s$$
$$r = \dfrac{s}{2s-1}$$
If $\, s=3$, then $r= \dfrac 35$ and $a = -\dfrac{1}{rs}=-\dfrac 59$.
if $\,s=4$, then $r= \dfrac 47$ and  $a = -\dfrac{1}{rs}=-\dfrac{7}{16}$.
So $-\dfrac59\le a<-\dfrac7{16}$.
