Eq. ckt of fig 5.b.1 is shown as

5.b.1

5.b.2

As Icr=0 so Vcr=0. Hence resistor can be replaced with short ckt .

As there is a direct connection between drain and gate ∴ VDS=VGS

From output ckt; VDS=VDD-IDRD

$where, \\ I_D=I_{DSS}\left(1-\dfrac{V_{GS}}{V_P}\right)^2\\ \therefore V_{DS}=V_{DD}-I_{DSS}\left(1-\dfrac{V_{DS}}{V_P}\right)^2RD[V_{GS}=V_{DS}]\\ \therefore V_{DS}=20-8\left(1+\dfrac{V_{DS}}{4}\right)^24\\ \therefore V_{DS}=20-2\left(1+\dfrac{{V_{DS}}^2}{16}+\dfrac{2V_{DS}}{4}\right)$

$16 V_{DS}=320-2[16+{V_{DS}}^2+8V_{DS}]\\ 16V_{DS}=320-32-2V_{DS}-288=0\\ {V_{DS}}^2+16V_{DS}-144=0\\ V_{DS}=\dfrac{-16V_{DS}\pm\sqrt{256+576}}{2}\\ =\dfrac{-16+25.13}{2}=4.56\\ V_{DS}=V_{GS}=4.56V\\ I_{DQ}=8\left(1+\dfrac{20.8}{4}\right)=6.2\times8mA\\ =49.7mA$