Let the two resistances be R1 and R2 When they are connected in series, equivalent resistance, Rs=6Ω R1+R2=6 → (1) When they are connected in parallel, equivalent resistance, Rp=
4
3
Ω
1
Rp
=
1
R1
+
1
R2
⇒
1
R1
+
1
R2
=
3
4
→ (2) Solving Eq. (2), we get
R1+R2
R1R2
=
3
4
⇒R1R2=(R1+R2)×
4
3
⇒R1R2=6×
4
3
⇒R,R2=2×4=8Ω (Using Eq.(1)) ⇒R1R2=2×4=8Ω → (3) From Eq. (1), we have R2=6−R1
⇒R1(6−R1)=8⇒6R1−R12=8⇒R12−6R1+8=0 R1=
6±√36−32
2
=
6±√4
2
=
6±2
2
=4 or 2
Using Eq. (3), we get R2=2 or 4 Therefore, the two resistances are 4 Ω and 2 Ω.