Two resistances when connected in parallel give resultant value of `2Omega`, when connected in series the value becomes `9Omega`. Calculate the value of each resistance.

Explanation

Let `R_(1)andR_(2)` be the two resistances.
When they are in parallel, resultant resistance `R_(p)` is given by
`1/R_(p)=1/R_(1)+1/R_(2)`
`R_(p)=(R_(1)R_(2))/((R_(1)+R_(2)))`
`2=(R_(1)R_(2))/((R_(1)+R_(2)))`
`2[R_(1)+R_(2)]=R_(1)R_(2)`
When they are in series effective resistance
`R_(s)=R_(1)+R_(2)`
`9=R_(1)+R_(2)rArrR_(2)=9-R_(1)`
`2xx9=R_(1)R_(2)=R_(1)[9-R_(1)]=9R_(1)-R_(1)^(2)`.
`18=9R_(1)-R_(1)^(2)`
`R_(1)^(2)-9R_(1)+18=0`
Solving by using quadratic formula, we get
`R_(1)=((9pmsqrt(81-4xx1xx18))/2)" "R_(1)=((9pmsqrt9)/2)=((9pm3)/2)`
`R_(1)` = 3 ohm or 6 ohm
If `R_(1)` = 3 ohm, `R_(2)` = 6 ohm and vice versa.