Physics Question?

Question

I am struggling with this question and can't arrive at an answer. Any detailed guidance/help anyone could give would be greatly appreciated.

A light bulb is wired in series with a 132 ohm resistor, and they are connected across a 138.0 V source. The power delivered to the light bulb is 22.7 W. What are the two possible resistances of the light bulb?

lower value= ?
higher value= ?

I know that V=IR and P=IV and that my equivalent resistance is going to be R= 132 +x but past that I'm lost.

Answer 1

V = IR
138 = I(132 + R)
P = (I^2)R (P = IV = I*IR)
22.7 = I^2R
Solving for I,
R = 22.7/I^2
138 = 132I + 22.7/I
132I^2 - 138I + 22.7 = 0
I = (138 ± √(138^2 - 4*132*22.7))/264
I = (138 ± √7058.4)/264
I = (138 ± 84.01428)/264
I = 0.2044913, 0.8409632
Substituting back into R = 22.7/I^2
R = 542.8452, 32.09755

Answer 2

I always figure out what's conserved and go from there...

The current is the same through the bulb and R

Call the bulb, subscript 1
resistor : subscript 2

I1 = I2
P/V1 = V2/R2
V1 + V2 = V = 138 volts

You know P, R2

So, you have two equations in the two unknowns, V1, V2
Get those and then everything is known. Hmm, you may be right about two possible values.

You might end up with a quadratic when you combine your two equations into one.