Rearranging two almost identical implicit equations I'm aware this question may appear simple, but here it is:
How do you get from this:
$\frac{2h'+s}{\sqrt{h'}}=\frac{2h+s}{\sqrt{h}} $
To this:
$s=2\sqrt{hh'}$

$h'(2h+s)^2=h(2h' +s)^2 $
$h(4h'^2+4h's+s^2) = h'(4h^2+4hs+s^2)$
$4h'^2h + 4h'sh+s^2h=4h^2h'+4hsh'+s^2h'$
$4h'^2h+s^2h = 4h^2h' + s^2h'$
$4((h')^2h + h^2h')=s^2(h'-h)$
$s^2 = \frac{4(h')^2h + h^2h'}{h'-h}$
From here it went pear-shaped:
$s = \frac{2\sqrt{(h')^2h + h^2h'}}{\sqrt{h'-h}}$
$s = \frac{2\sqrt{(h')^2h + h^2h'}\sqrt{h'+h}}{\sqrt{h'-h}\sqrt{h'+h}}$
$s = \frac{2\sqrt{(h')^3h+2(hh')^2+h^3h'}}{\sqrt{(h')^2-h^2}}$
I couldn't see how I could get from here to the answer.
I've made several other attempts with different methods but they all had the same outcome.
Thank you for your help.
 A: at the LHS: $\frac{2h'+s}{\sqrt{h'}}=\frac{2h'}{\sqrt{h'}}+\frac{s}{\sqrt{h'}}=2\sqrt{h'}+\frac{s}{\sqrt{h'}}$
at the RHS: $\frac{2h+s}{\sqrt{h}}=\frac{2h}{\sqrt{h}}+\frac{s}{\sqrt{h}}=2\sqrt{h}+\frac{s}{\sqrt{h}}$
we left with $2\sqrt{h'}+\frac{s}{\sqrt{h'}}=2\sqrt{h}+\frac{s}{\sqrt{h}}$
now we can change thing sides like this:$$2\sqrt{h'}+\frac{s}{\sqrt{h'}}=2\sqrt{h}+\frac{s}{\sqrt{h}}\\\implies2\sqrt{h'}-2\sqrt{h}=\frac{s}{\sqrt{h}}-\frac{s}{\sqrt{h'}}\\\implies2(\sqrt{h'}-\sqrt{h})=s(\frac{1}{\sqrt{h}}-\frac{1}{\sqrt{h'}})$$
can you go from here?
A: Multiplying both sides by $\sqrt{hh'}$ results in:

 $$2h'\sqrt{h}+s\sqrt{h}=2h\sqrt{h'}+s\sqrt{h'}.$$

Rearranging to bring the $s$'s together and the other terms together gives

 $$s\sqrt{h}-s\sqrt{h'}=2h\sqrt{h'}-2h'\sqrt{h}.$$

Factoring and breaking $h$ into $\sqrt{h}\sqrt{h}$ gives 

 $$s(\sqrt{h}-\sqrt{h'})=2\sqrt{h}\sqrt{h}\sqrt{h'}-2\sqrt{h'}\sqrt{h'}\sqrt{h}=2\sqrt{h}\sqrt{h'}(\sqrt{h}-\sqrt{h'})$$

Assuming that $h\not=h'$, we can divide and get the desired result.
A: WLOG $\sqrt h=H,\sqrt{h'}=H'$
Method$\#1:$ $$\implies\dfrac{nH^2+s}H=\dfrac{nH'^2+s}{H'}=\dfrac{nH^2+s-(nH'^2+s)}{H-H'}=n(H+H')\text{ if } H-H'\ne0$$
$$\implies nH^2+s=Hn(H+H')\iff s=?$$
Method$\#2:$
$\dfrac{nH^2+s}H=\dfrac{nH'^2+s}{H'}=K$(say)
So, $H,H'$ are the roots of $$nt^2-tK+s=0\implies HH'=\dfrac sn$$
