The current in a series RLC circuit varies with frequency as shown in the graph
Current is calculated as
At critical frequency current is
Combining these two we get
at resonance impedance of the circuit is simply R. Now writing Io =V/R and substituting value of Z
Cancelling V from both sides and taking reciprocal on both sides
To get rid of j (equivalent i of imaginary number), write the right hand side expression in terms of it's magnitude form
square on both sides
and simplify
Now square root both sides
Substitute in value for inductive and capacitive impedance in terms of omega,L,C
Further simplify
This is a quadratic equation in ω.
Because of +- sign there are two different
quadratic equations. So considering one at a time
Now calculating roots of the above quadratic equation
Now here there are two possible roots. Since frequency cannot be negative so
considering positive root.
lets call this ω1 (lower cut-off frequency)
Now considering the second quadratic
equation
calculating roots
Now just for observation if you factor out R/4L
form inside the squared root, we can see that the term is actually greater than R/2L term, so this term cannot be negative as will make our frequency negative which is fundamentally wrong.
So choosing the positive root
Lets call this ω2 (higher cut-off frequency)
We can write both the roots together as