If  p, q, r and s are non zero numbers such that ‘r’ and ‘s’ are roots of  $x^2+px+q=0$  and  $p,q$  are the roots of  $x^2+rx+s=0$ , then the value of  $(4p+q+2r+s)$  is .

$x^2+px+2=0{<}_s^r$  and  $x^2+rx+s=0{<}_q^p$
$r+s=-p$  ….. (1)
$rs=q$  …… (2) 
$p+q=-r$  ….. (3)
$pq=s$  ……… (4)
$\left(2\right)+\left(4\right)\Rightarrow pqrs=qs\Rightarrow pr=1$  ….. (5)
 $\left(1\right)\times \left(3\right)\Rightarrow p+q+r+s=-p-r$
$\Rightarrow 2(p+r)+q+s=0$  …….. (6)
Now p is a root of  $x^2+rx+s=0\Rightarrow p^2+rp+s=0$
and r is a root of  $x^2+px+q=0\Rightarrow r^2+pr+q=0$
adding those two
 $\Rightarrow {(p+r)}^2-2(p+r)=0\Rightarrow p+r=0,p+r=2$
also $pr=1\Rightarrow p=1,r=1$
From (4)  $pq=s\Rightarrow q=s$
 $\therefore q+s=-2(p+r)$
 $\Rightarrow q+s=-4\Rightarrow q=s=-2$
Required value is   $4\times 1-2+2-2=2$