Quadratic equation and trig If the quadratic equation $ax^2+bx+c=0$ has equal roots where $a, b$ and $c$ denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of 
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$
Just some hints will do. 

Let $s=\frac ab,t=\frac cb$. Then, we have
$$b^2-4ac=0\Rightarrow 4st=1.$$
So, we have
$$\begin{align}&a+b\gt c,\quad b+c\gt a,\quad c+a\gt b\\&\iff s+1\gt t,\quad 1+t\gt s,\quad t+s\gt 1\\&\iff s+1\gt \frac{1}{4s},\quad 1+\frac{1}{4s}\gt s,\quad \frac{1}{4s}+s\gt 1\\&\iff \frac{\sqrt 2-1}{2}\lt s\lt \frac 12\quad\text{or}\quad \frac 12\lt s\lt \frac{1+\sqrt 2}{2}\end{align}$$
Now note that we can write
$$\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}=\frac ac+\frac ca=\frac{a^2+c^2}{ac}=\frac{a^2+c^2}{\frac{b^2}{4}}=4s^2+4t^2=4s^2+4\left(\frac{1}{4s}\right)^2.$$
Let $f(s)=4s^2+\frac{1}{4s^2}$. Then, 
$$f'(s)=\frac{16s^4-1}{2s^3}.$$
Hence, $f(s)$ is increasing only when $s\ge \frac 12$. Thus, we have
$$2= f\left(\frac 12\right)\lt f(s)\lt f\left(\frac{\pm 1+\sqrt 2}{2}\right)=6.$$
Thus, the answer is $3+4+5=12$. (note that $f(s)\not=2$.)
