Radical Simplification

Note that $5+2\sqrt{6}=\sqrt{3}^2+2\sqrt{3}\sqrt{2}+\sqrt{2}^2=(\sqrt{3}+\sqrt{2})^2.$ Since $\sqrt{3}+\sqrt{2}>0,$ we have $\sqrt{5+2\sqrt{6}}=\sqrt{(\sqrt{3}+\sqrt{2})^2}=\sqrt{3}+\sqrt{2}.$ Hence, $a=3,$ $b=2$ and $a-b=\boxed{1}.$

square LHS & RHS then we get 5 + 2√6 = a+b+2√ab ( (a+b)^2 = a^2 + b^2 + 2ab). On comparing both the sides we see that a+b = 5 and ab = 6. hence we conclude that a , b = 3 , 2 respectively because a>b. therefore a-b = 3-2 = 1

Squaring both sides,we get: $\color{#D61F06}{5+2\sqrt6=a+b+2\sqrt{ab}\\ \rightarrow a+b=5\\ \rightarrow 2\sqrt{ab}=2\sqrt{6} \rightarrow ab=6}$ We see that $\color{#20A900}{a=3,b=2}$ is a solution so $\color{#3D99F6}{\sqrt{a}+\sqrt{b}=\sqrt{3}+\sqrt2\rightarrow a-b=3-2=\boxed{1}}$

We are given the equation

$\sqrt{5+2\sqrt{6}} = \sqrt{a} + \sqrt{b}$

Where $a > b$

First we square both sides to get

$5 + 2\sqrt{6} = \sqrt{a}^2 + 2\sqrt{a}\sqrt{b} + \sqrt{b}^2$

$5 + 2\sqrt{6} = a+b+2\sqrt{ab}$

From this we can gather that

$a + b = 5$

$ab = 6$

We can use these equation to find $a$ and $b$

$b = 5 - a$

$a(5 - a) = 6$

$5a - a^2 = 6$

$-a^2 + 5a - 6 = 0$

$a^2 - 5a + 6 = 0$

Now we can use the quadratic formula for $a$

$a = \frac{-(-5)\pm\sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$

$a = \frac{5\pm 1}{2}$

$a = 3, 2$

Alternately you can also factorise.

We know $a>b$ and that $a+b = 5$ so $a$ can't be equal to $2$ as that would suggest that $b = 3$ which is incorrect. This means that $a = 3$ and $b = 2$ so

$a-b = 3-2 = 1$

sqrt(5+(2sqrt(6)))=sqrt(5+6)=sqrt(5)+sqrt(6) but a > b so sqrt(5+6)=sqrt(6)+sqrt(5) so the answer is 6-5=1

5+2Sqrt(6)=a+b+2Sqrt(ab)
So a+b=5 ab=6
a=3 b=2 --> a-b=1