Quadratic fun-1

Let $x^2+(2a+1)x+(a^2+2)=(x-r)(x-2r)=x^2+(-3r)x+2r^2$ , where $r, 2r$ are our desired roots. From this we have our simultaneous equations:

$2a+1=-3r$

$\Rightarrow r = -\frac{2a+1}{3} - (1)$

$a^2+2=2r^2 - (2)$

Substituting (1) into (2):

$a^2+2=2(-\frac{2a+1}{3})^2=\frac{8a^2+8a+2}{9}$

$\Rightarrow 9(a^2+2)=9a^2+18=8a^2+8a+2 \Rightarrow 8a=16 \Rightarrow a=2$ .

This is the only value of $a$ that we have acquired, hence the sum of all such values $a$ that fulfil the problem's conditions is thus $\boxed{2}$

let r be one root and 2r is the other root sum of roots=-2a-1=3r (eq.1) prod of roots=a^2+2=2r^2 (eq2) solving the equations will give (substitute the value of r in eq1 to eq2)

a=4