So, I've been stuck with this equation for a whole day now and can't seem to get to the answer. I tried in Mathematica, solved but I can't solve in manually. We have
$a' = -\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\left[1+\dfrac{4}{\lambda^2}a(1-a)\right]}$
which is to be solved by $\dfrac{d}{da}a'(1-a)=0$ to get $\lambda^2=\dfrac{(1-a)(4a-1)^2}{1-3a}$. I solved it manually and for verification checked on Mathematica but here is from where I cannot reproduce the answer
$\lambda^2-\lambda^2\sqrt{1+\dfrac{4a}{\lambda^2}(1-a)}=2(1-4a)(1-a)$
I get stuck on the $\lambda$ with the radical sign. For information it is from Aerodynamic Design of Horizontal Axis Wind Turbine Blades By Asress and Gerawork Eq 38/39
Using $\Lambda=\lambda^2$ $$a' = -\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\left[\dfrac{4}{\Lambda}a(1-a)+1\right]}$$ $$\dfrac{d}{da}\left(a'(1-a)\right)=\frac{(1-a) \left(\frac{4 (1-a)}{\Lambda }-\frac{4 a}{\Lambda }\right)}{4 \sqrt{\frac{4 (1-a) a}{\Lambda }+1}}-\frac{1}{2} \sqrt{\frac{4 (1-a) a}{\Lambda }+1}+\frac{1}{2}$$ Use common denominator to get $$\dfrac{d}{da}\left(a'(1-a)\right)=\frac{\Lambda \left(\sqrt{1-\frac{4 (a-1) a}{\Lambda }}-1\right)+2 a (4 a-5)+2}{2 \Lambda \sqrt{1-\frac{4 (a-1) a}{\Lambda }}}$$ Since you want this to be $0$, then $$\Lambda \left(\sqrt{1-\frac{4 (a-1) a}{\Lambda }}-1\right)=2a(5-4a)-2$$ Just square it to get the result.