Show that every solution tends to zero as $x→∞$ if $a_1>0$ for $y''+a_1y'+a_2y=0$. Please help me with the following problem:

Consider the equation $y'' + a_1y' + a_2y = 0$, where the constants $a_1, a_2$ are real. Suppose $α + iβ$ is a complex root of the characteristic polynomial, where $α, β$ are real, $β≠0$.

Show that every solution tends to zero as $x→∞$ if $a_1>0$.

My solution:
$y'' + a_1y' + a_2y = 0$
$r^2 + a_1r + a_2 = 0$
Using quadratic equation,
$x = \frac{-a_1}{2} ±\frac{\sqrt{a_1^2-4a_2}}{2}$
$\varphi(x) = c_1e^{(\frac{-a_1}{2}+\frac{\sqrt{a_1^2-4a_2}}{2})x} + c_2e^{(\frac{-a_1}{2}-\frac{\sqrt{a_1^2-4a_2}}{2})x}$

I do not know where to go from here. Please help!
 A: If $a_1^2<4a_2$, then $\sqrt{a_1^2-4a_2}=i\sqrt{4a_2-a_1^2}$ is imaginary and 
$$\begin{align}
\lim_{x\to \infty}e^{\frac12\left(-a_1\pm\sqrt{a_1^2-4a_2}\right)x}&=\lim_{x\to \infty}e^{-a_1x/2}e^{\pm\frac i2\sqrt{4a_2-a_1^2}}\\\\
&=\lim_{x\to \infty}e^{-a_1x/2}\left(\cos\left(\frac12\sqrt{4a_2-a_1^2}x\right)\pm i\sin\left(\frac12\sqrt{4a_2-a_1^2}\right)x\right)\\\\
&= 0
\end{align}$$

If $a_1^2\ge 4a_2$, and $a_2>0$, then $0\le \sqrt{a_1^2-4a_2}<a_1$.  
Hence $-a_1+\sqrt{a_1^2-4a_2}< 0$ and $e^{\frac12\left(-a_1\pm\sqrt{a_1^2-4a_2}\right)x}\to 0$ as $x\to \infty$.
A: Split the exponentials into a product. For $a_1 > 0$ $e^{-a_1 x/2} \to 0$ as $x \to \infty$.
A: The characteristic polynomial's roots have sum $-a_1<0$ and product $a_2>0$, so are both negative. The solution is therefore exponentially suppressed and vanishes as $x\to\infty$. (Note that if $a_2<0$ we would instead have a positive root, giving a solution that diverges for large $x$, while if $a_2=0$ we have $y_1=A+Be^{-a_1 y}$ for constants $A,\,B$, agreeing with the required result iff $A=0$.)
