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.
When is the given function positive
0
$\begingroup$
algebra-precalculus
quadratics
-
0This 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. – 2012-10-22
1 Answers
3
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 $x$.
-
0thank you!! that solved everything – 2012-10-22