Basic ac circuit analysis
"notice that Xl is not written as jXl"


You've answered your own question. The problem is that \$X_L\$ is not written as \$j X_L\$. :-)

While it also has the dimension &Omega;, just like impedance, \$X\$ is the symbol for reactance, not impedance, that would be \$Z\$.

\$ Z_L = R + X_L = R + j \omega L\$

The factor \$j\$ indicates a 90° phase shift with respect to the resistance which does not have a phase shift. For an inductance the phase shift is positive, for a capacitance it's negative, since

\$ Z_C = R + X_C = R + \dfrac{1}{j \omega C} = R - j \dfrac{1}{\omega C}\$

It's because of the 90° angle between resistance and reactance that you can't add them arithmetically, like 5 &Omega; + 12 &Omega; = 17 &Omega;. \$X\$ and \$R\$ create a right angled triangle, so you have to apply Pythagoras to find the magnitude of the impedance: 

\$ |Z_L| = \sqrt{|R|^2 + |X_L|^2} \$

so for \$R\$ = 5 &Omega; and \$X_L\$ = \$j\$ 12 &Omega; we get a magnitude of 13 &Omega; instead of 17 &Omega;. 

Right, back to your calculations. The reason it seems to go in the right direction, even without \$j\$ is that you applied Pythagoras implicitely in the first line already; you wrote 

\$ V_S^2 = (V_1 + V_R)^2 + V_L^2 \$

and not 

\$ V_S^2 = (V_1 + V_R + V_L)^2 \$

or simply

\$ |V_S| = |V_1 + V_R + V_L| \$

That's why you won't see \$j\$ anymore; you got rid of it at the start.