How to do DC Analysis for a Common Source Biasing Circuit using NMOS

I am having a lot of trouble with this specific problem. I want to find Vg, Vd, Vs, and Id for this specific common source biasing circuit that has an NMOS. I have these values given: Vdd = 9V, k/2 = 2.9, Vt 1.55 V

So far I have these equations (assuming active/saturation mode)

Vg = 9V * (10k/(10k+10k)) = 4.5 V (I used voltage divider here)

Then using the equation for active/saturation mode:

Id = k/2(Vgs - Vt)^2 = 2.9(Vgs - 1.55)^2

Last I tried using KVL for these equations:

GS loop: 9V - Vgs - 1k*Id = 0

DS loop: 9V - 2kId - Vds - 1kId = 0

When I try solving equations with these loop equations, I get very stuck. I get incorrect/differing answers depending on what variable I substitute. Can anyone help me solve this problem?

The nature of these MOSFET DC analysis problems is that you end up with a quadratic equation and you have to pick the solution which satisfies the assumption you made at the beginning.

You assumed saturation and I will do the same. Let's see if there is a solution which satisfies our assumption.

First, for saturation, in the end, we have to verify that two conditions are met: $$V_{GS} > V_t$$ $$V_{DS}\geq V_{GS}-V_t$$

So, for saturation, as you have it in your post: $$I_D=k(V_{GS}-V_t)^2$$I am absorbing the $\frac{1}{2}$ into $k$.

What I will do now is replace $V_{GS}$ with some equation that includes $I_D$. $$V_{GS} = V_G-V_S=V_G-I_DR_2$$ If we now plug this into our $I_D$ equation, we have $$I_D=k(V_G-I_DR_2-V_t)^2$$ The math gets a little messy, I hope I didn't miss anything. But if you expand the quadratic term and collect the $I_D$ terms, you should end up with something like this: $$I_{D}^2(kR_2^2)+I_{D}(2kR_2V_t-2kV_GR_2-1)+(kV_G^2-2kV_GVt+kV_t^2)=0$$

I used MATLAB to solve for $I_D$ and I get the following roots: $$I_{D1}=3mA$$ $$I_{D2}=2.9mA$$

For the first solution, $V_{GS}=1.5V<V_t$. You can rule out this solution since it won't satisfy our assumption of saturation. It would mean the MOSFET is cut off.

The second solution gives $V_{GS}=1.6V>V_t$. That's a good start, but we still need to satisfy the second condition of $$V_{DS}\geq V_{GS}-V_t$$

$$V_{DS}=V_{DD}-I_D(R_1+R_2)=0.3V$$ $$V_{GS}-V_t =0.05V$$ And $0.3V\geq 0.05V$. So the second condition is met and the MOSFET is in saturation.

I am not sure if I used the right units for $k$. Usually this is expressed in terms of $\frac{mA}{V^2}$ or $\frac{\mu A}{V^2}$. So I used $\frac{A}{V^2}$ since no units were given. This could change the region of operation for the MOSFET (could be in triode, for example).

All in all, that is the approach that I would take. I hope it helps!