Question

12V battery in series with 3 resistros.
Given: R1=1 ohm
Voltage drop accross R2=4V
Power accross R3=12V.
What is current in the circuit?

Answer 1

I^2 - 8 I +12 = 0 is the correct quadratic equation
I = - (-8/2) [1 +- sqrt(1 - 12/(8/2)^2)]
I = 4 [1 +- sqrt((1- 12/16)]
I = 4 [1 +- sqrt{1 - 3/4)]
I = 4 [1 +- 1/2]
I = 6, 2
The choice of the two answers would be dependent on the required voltage across the 1 ohm resistor which is dependent on R2 and R3