∣x−2∣2+2∣x−2∣−∣x−2∣−2=0⇒(∣x−2∣+2)(∣x−2∣−1)=0⇒∣x−2∣=1⇒x=2±1=3,1⇒ sum of square of roots =9+1=10x2−2∣x−3∣−5=0 Case-I x−3≥0⇒x2−2x+1=0⇒(x−1)2=0⇒x=1
But x≥3
⇒x∈ϕ
Case-II x−3<0
x2+2x−11=0,D>0⇒ Real & distinct roots
f(x)=x2+2x−11
f(3)>0,2a−p=−1<3
⇒ both roots <3, both roots acceptable
Sum of square of roots =(α+β)2−2αβ
$\begin{aligned}
& =4+22=26 \
& \Rightarrow \text { Final sum }=10+26=36
\end{aligned}$