Find all points $(a,b)$ for which all roots of $ax^2 + 2bx + 4a = 0$ are higher than 1.

Question

Find all points $(a,b)$ for which all roots of $ax^2 + 2bx + 4a = 0$ are higher than 1.

I found out that roots are of form $\frac{-b + \sqrt{b^2 - 4a^2}}{a}$ and $\frac{-b - \sqrt{b^2 - 4a^2}}{a}$, but I have no idea what to do next.

Answer 1

If $ax^2 + 2bx + 4a = 0 $, dividing by $a$ we get $x^2+(2b/a)x = -4 $.

We can not have $a=0$, because this becomes $2bx=0$ which root $0$ which is too small.

To simplify, let $c = b/a$. We then have $x^2+2cx = -4 $.

Adding $c^2$, $-4+c^2 =x^2+2cx+c^2 =(x+c)^2 $ or $x =-c\pm\sqrt{-4+c^2} $.

For this to have real roots, then $c^2 \ge 4$ or $|c| \ge 2$.

If $c \ge 0$, then $-c-\sqrt{-4+c^2} < 0 $, so one root is negative, which is out.

Therefore, we must have $c < 0$. Since we need $|c| \ge 2$, we must have $c \le -2$.

Let $c = -d$, so we are looking at $d \pm \sqrt{d^2-4}$. The smaller of these is $d - \sqrt{d^2-4}$, so we want $d - \sqrt{d^2-4}> 1$, or $d-1 > \sqrt{d^2-4} $.

Squaring, since $d-1 > 1$, $d^2-2d+1 > d^2-4 $ or $2d < 5$ or $d < \frac52$.

Therefore $2 \le d < \frac52$ or $-2 \ge c > -\frac52$.

Putting the definition of $c$, this becomes $-2 \ge \frac{b}{a} > -\frac52$.