When is the given function positive

I have to find the value of $x$ for which the given function is positive \begin{align} \alpha +\beta x + \sqrt{ax^2+bx+c} \end{align} I know that $ax^2+bx+c$ is always positive. Given conditions, $x>0$, $\beta<0$ and $a,b,c>0$ and $\alpha>0$. Rest of constants are real in general.

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If you don't know that $a,b\geq0$, then how do you know that $ax^2+bx+c>0$ for all $x$? –  Dennis Gulko Oct 22 '12 at 8:07
I agree with @Dennis. How do we know $ax^2+bx+c>0$? Unless your problem, previously assumed that as an initial assumption. –  Babak S. Oct 22 '12 at 8:47
This happens due to the way the problem was formulated.I don't mind if you assume $a>0$ and $b>0$. I have edited the question. –  dineshdileep Oct 22 '12 at 8:50

Outline of the solution: Since $\beta<0$, let's write $\alpha-\beta x$ with $\beta>0$: $$\alpha -\beta x + \sqrt{ax^2+bx+c}>0 \hspace{5pt}\Rightarrow\hspace{5pt} \sqrt{ax^2+bx+c}>\beta x-\alpha$$ Now we have two cases:
I) $x\geq\frac\alpha\beta$. Then we can square both sides: $$ax^2+bx+c>\beta^2x^2-2\alpha\beta x+\alpha^2\hspace{5pt}\Rightarrow\hspace{5pt} (a-\beta^2)x^2+(b+2\alpha\beta)x+c-\alpha^2>0$$ Now you need to assume that either $a>\beta^2$, $a<\beta^2$ or $a=\beta^2$. According to those you can find the range of $x$ and combine it with $x\geq\frac\alpha\beta$.
II) $x<\frac\alpha\beta$. Then the RHS is negative. Since the LHS is always non-negative, the inequality holds for all such $x$.