Question

two resistance when connected in parallel give resultant value of 2 ohm;when connected in series the value becomes 9 ohm.calculate the value of each resistance.
plz answer fast

Answer 1

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

When they are in series effective resistance 

Rs = R1 + R2
9 = R1 + R2 ⇒ R2 = 9 - R1

2 x 9 = R1 R2
          = R1 (9 - R1)
          = 9R1 - R12.
18 = 9R1 - R12.
R12 - 9R1 + 18 = 0

Solve this quadratic equation. You will get 

R1 = (9 ± √(81 - 4 x 1 x 18))/2
      = (9 ± √9)/2
      =  (9 ± 3)/2
R1 = 3 ohm or 6 ohm

If R1 = 3 ohm, R2 = 6 ohm and vice versa.