What this shows is that if $a$ and $b$ are going to satisfy $\sqrt{a}-\sqrt{b}=\sqrt{a-b}$ then $a$ must be equal to $b$. In fact part of the solution has been missed because in the 7th line we divided by $\sqrt{b}$ which assumed $b \neq 0$, so in the fact the possibilites are that $a=b$ or $b=0$. Conversely we can check that if $a=b$ or $b=0$
then $a$ and $b$ satisfy$\sqrt{a}-\sqrt{b}=\sqrt{a-b}$. So in summary $\sqrt{a}-\sqrt{b} =\sqrt{a-b}$ if and only if $a=b$ or $b=0$.
Sergio from King's College of Alicante has found a solution to i) $\sqrt{5+2 \sqrt{6}} = \sqrt{a}+\sqrt{b}$
In the first place I decided to square everything to get rid of some square roots:
$ 5 + 2 \sqrt{6} = a + b + 2 \sqrt{ab}$
Now I looked carefully and I noticed that I probably had found two equations, so I now needed to solve simultaneous equations!
$5=a+b$
$6=ab$
This means that
$6=b(5-b)$
$0=b^2-5b+6$
I solved it by factorising, although you could have also used the quadratic equation, which gave me that $b$, and also $a$ when I substituted in, could be either 3 or 2.
Thomas from BHASVIC Sixth Form College Brighton has solved the remaining parts:
a) Squaring both sides of the equation we simply get $a \times b = ab$ which is clearly true for all non-negative $a$ and $b$.
b) Again, squaring both sides gives $\frac{a}{b} = \frac{a}{b}$, which again is true for all non-negative $a$ and $b>0$.
c) The largest value of $b$ occurs at $a=0$, when $b=23-6\sqrt{6-4\sqrt{2}}$. As $a$ increases, $b$ decreases until it reaches zero, and $a=23-6\sqrt{6-4\sqrt{2}}$. Squaring gives the solution set $a=(\sqrt{23-6\sqrt{6-4\sqrt{2}}}-\sqrt{b})^2$ for $0 \leq b \leq 23-6\sqrt{6-4\sqrt{2}}$.
d) Squaring both sides gives $a^2b=ab \implies ab(a-1) = 0$. So have solutions $a=0,1$ for all non-negative $b$, and $b=0$ for all non-negative $a$.
e) Divide the top and bottom of the fraction $\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$ by $\sqrt{a}$. Then multiply through by denominator and rearrange to isolate $\sqrt{a}$:
$$\sqrt{a}=\frac{\sqrt{b}}{\sqrt{b}-1}$$
The LHS is positive, so must have denominator $\sqrt{b}-1>0$, so $b>1$. So solutions are given by
$$a = \frac{b}{(\sqrt{b}-1)^2}$$
for $b>1$.
g) Squaring both sides of the equation gives $$a+b+2\sqrt{ab}=a+b+\sqrt{4ab}$$ which is clearly true for all non-negative $a$ and $b$.
h) Take the RHS to the LHS of the equation and factorise out $\sqrt{a}+b$, so that
$$(\sqrt{a}+b)(\frac{c-d^2}{\sqrt{c}+d})=0$$
Then have the solution sets:
$a=b=0$ for any $c$ and $d$ non-negative not both equal to zero
or
any non-negative $a,b$ and all positive $c=d^2$.