Just Apply Vieta's!

Let's assume the roots to be $\alpha$ and $2 \alpha$

Using Vieta's we have:-

$\alpha+2\alpha= -(2a+1)$

And, $2 \alpha ^2=a^2+2$

From the 1st Relation, We get:- $\alpha=\dfrac {-(2a+1)}{3}$

Putting this value into the second equation, we have:-

$2(2a+1)^2=9(a^2+2)$

$= a^2-8a+16=0$

We get the unique value of $a=4$

Hence, The roots of the equation are:- $x^2+9x+18=0$

$(x+6)(x+3)=0$

$x=-6 \quad or -3$ . And , One of the roots of the equation is twice the other. :D

Answer= $\boxed 4$