Derivation of width of depletion layer in the pn-junction

[tcsv018]

Hello,

I read a derivation for the width of the depletion region $W$ in "SEMICONDUCTOR DEVICE FUNDAMENTALS" by Robert F. Pierret in which at one point it says:
http://imageshack.com/i/ipbgKsK9p
Here again for better readability:

$x_N = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_A}{N_D\cdot(N_A+N_D)}V_{bi}}$
$x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D}{N_A\cdot(N_A+N_D)}V_{bi}}$
$W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D+N_A}{N_A\cdot N_D}V_{bi}}$

Which is confusing to me as I would expect the same containing:
$W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D^2+N_A^2}{N_A\cdot N_D\cdot(N_A+N_D)}V_{bi}}$This same outcome though is found on various places in the internet.

Does anyone know what I am missing?

Kind regards,

Name

Symbols:
$x_N$ length of n-doped region
$x_P$ length of p-doped region
$N_A$density of acceptors
$N_D$density of donors
$V_{bi}$ builtin potential

All the others are known constants

 

[mfb]

$\sqrt{x}+\sqrt{y} \neq \sqrt{x+y}$

Ignoring the common prefactors,
$$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
$$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
$$W = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(\sqrt{N_A^2} + \sqrt{N_D^2}\right) = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(N_A + N_D\right) \\= \sqrt{\frac{(N_A+N_D)^2}{N_D N_A (N_A+N_D)}} = \sqrt{\frac{N_A+N_D}{N_D N_A}}$$

 

[tcsv018]

Thanks,
Shame on me