Given circuits are Circuit I,
Circuit II,
Clearly, both circuits are different after sections AB and CD.
So, resistances of two circuits are same, if RAB=RCD
So, R+X=R+6R+R+X6R(R+X) ⇒X=7R+X6R(R+X) ⇒7RX+X2=6X2+6RX ⇒X2+RX−6R2=0
From sridharacharya formula, we have ⇒X=2(1)−R±R2−4(1)(−6R2) =2−R±25R2
or X=2−R±5R ∴X=2R