Question

When two resistor of resistance R1 anf R2 are connected in parallel the net resistance is 3ohm, when connected in series, its value is 16ohm, calculate the values of R1 and R2

Answer 1

Let R1 and R2 be the two resistances. When they are in parallel, resultant resistance Rp is given by

1/Rp = 1/R1 + 1/R2
Rp = R1R2/(R1 + R2)
3 = R1R2/(R1 + R2)
3(R1 + R2) = R1R2

When they are in series effective resistance 

Rs = R1 + R2
16 = R1 + R2 ⇒ R2 = 16 - R1

3 x 16 = R1 R2
          = R1 (16 - R1)
          = 16R1 - R12.
48 = 16R1 - R12.
R12 - 16R1 + 48 = 0

Solve this quadratic equation. You will get

R1 = (16 ± √(256 - 4 x 1 x 48))/2
      = (16 ± √64)/2
      =  (16 ± 8)/2
R1 = 12 ohm or 4 ohm

If R1 = 12 ohm, R2 = 4 ohm and vice versa.
I hope it helped you