If the roots of the equation $ax^2-2bx+c=0$ are imaginary,then find the number of real roots of the equation $4e^x+(a+c)^2(x^3+x)=4b^2x$.
The only information i'm able to interpret is$-$
- Since the roots of $ax^2-2bx+c=0$ are imaginary,therefore $b^2<ac$.
- from the Descartes' rule $a,c<0$ and $b>0$
how to move further?
If $a$, $b$, and $c$ are real numbers, then the function $f$ has exactly one real root, where $$f(x):=4\,\exp(x)+(a+c)^2\,\left(x^3+x\right)-4b^2\,x\text{ for all }x\in\mathbb{R}\,.$$ To show this, observe that $$\begin{align}f'(x)&=4\,\exp(x)+(a+c)^2\,\left(3x^2+1\right)-4b^2\\&>(a+c)^2-4b^2>(a+c)^2-4ac=(a-c)^2\geq0\text{ for all }x\in\mathbb{R}\,,\end{align}$$ where we use $b^2<ac$ (from the fact that $a\,x^2-2b\,x+c=0$ has no real solutions). This proves that $f$ cannot have more than one real root. Because $f$ is continuous, as well as $$\lim_{x\to+\infty}\,f(x)=+\infty\text{ and }\lim_{x\to-\infty}\,f(x)=-\infty\,,$$ we conclude that a real root must exists, whence it is the only one.