Given area of square $= 9+6\sqrt{2}$ Without calculator show its length in form of $(\sqrt{ c}+\sqrt{ d})$ $\sqrt{9+6\sqrt{2}}$  to find length 
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$. 
 A: Hint:
$$9+6\sqrt{2}=6+6\sqrt{2}+3=(\sqrt{6}+\sqrt{3})^2$$
A: Let the side of the square be $(\sqrt c+\sqrt d)$ then its area is given as $$(\sqrt c+\sqrt d)^2=9+6\sqrt 2\tag 1$$
$$c+d+2\sqrt{cd}=9+6\sqrt 2$$ by comparing the corresponding  rational & irrational parts, we get $$c+d=9\tag 2$$
$$2\sqrt{cd}=6\sqrt 2\tag 3$$
Now, we know $$(\sqrt c-\sqrt d)^2=(\sqrt c+\sqrt d)^2-2\sqrt{cd}$$$$=(9+6\sqrt 2)-2(6\sqrt 2)$$$$(\sqrt c-\sqrt d)^2=9-6\sqrt 2\tag 4$$ multiplying (1) & (4) we get $$(\sqrt c+\sqrt d)^2(\sqrt c-\sqrt d)^2=(9+6\sqrt 2)(9-6\sqrt 2)$$
$$(c-d)^2=9\implies c-d=\pm 3$$ 1. taking positive sign, we get $c-d=3\tag 5$ solving (2) & (5), we get $$\color{red}{c=6, \ d=3}$$
2. taking negative sign, we get $c-d=-3\tag 6$ solving (2) & (6), we get $$\color{red}{c=3, \ d=6}$$ Both the above cases give the same result. Hence, the side of the square is $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 6+\sqrt 3}$$
or $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 3+\sqrt 6}$$
A: Notice, let the side of the square be $a$ then its area is given as $$a^2=9+6\sqrt 2$$
$$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 3)\sqrt 2$$
$$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 2)(\sqrt 3)$$
$$a^2=6+3+2\sqrt{6}\sqrt 3$$ $$a^2=(\sqrt 6)^2+(\sqrt 3)^2+2(\sqrt{6})(\sqrt 3)$$
Using $a^2+b^2+2ab=(a+b)^2$, we get $$a^2=(\sqrt 6+\sqrt 3)^2$$
hence, side of square $$a=\color{red}{\sqrt6+\sqrt 3}$$
